剑指 Offer - 学习

1 - 剑指 Offer 09. 用两个栈实现队列

class CQueue {
    Stack<Integer> a, b;
    // a : insert
    // b : delete
    public CQueue() {
        a = new Stack<>();
        b = new Stack<>();
    }
    public void appendTail(int value) {
        a.push(value);
    }
    public int deleteHead() {
        if (b.empty())
        {
            while (!a.empty())
                b.push(a.pop());
        }
        return (b.empty()) ? -1 : b.pop();
    }
}

2 - 剑指 Offer 30. 包含min函数的栈

class MinStack {
    /** initialize your data structure here. */
    Stack<Integer> a, b;
    public MinStack() {
        a = new Stack<>();
        b = new Stack<>();
    }
    public void push(int x) {
        a.push(x);
        if (b.empty() || x <= b.peek())
            b.push(x);
    }
    public void pop() {
        if (a.pop().equals(b.peek()))
            b.pop();
    }
    public int top() {
        return a.peek();
    }
    public int min() {
        return b.peek();
    }
}

3 - 剑指 Offer 06. 从尾到头打印链表

class Solution {
    public int[] reversePrint(ListNode head) {
        Stack<Integer> a = new Stack<>();
        int n = 0;
        for (; head != null; head = head.next)
        {
            a.push(head.val);
            n ++;
        }
        int[] b = new int[n];
        int m = 0;
        for (; !a.empty(); )
        {
            b[m ++] = a.pop();
        }
        return b;
    }
}

4 - 剑指 Offer 24. 反转链表

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev, cur, next;
        for (prev = null, cur = head; cur != null; prev = cur, cur = next)
        {
            next = cur.next;
            cur.next = prev;
        }
        return prev;
    }
}

5 - 剑指 Offer 35. 复杂链表的复制

class Solution {
    public Node copyRandomList(Node head) {
        HashMap<Node, Node> a = new HashMap<>();
        Node cur, b;
        for (cur = head; cur != null; cur = cur.next)
        {
            b = new Node(cur.val);
            a.put(cur, b);
        }
        for (cur = head; cur != null; cur = cur.next)
        {
            Node c = a.get(cur);
            c.next = a.get(cur.next);
            c.random = a.get(cur.random);
        }
        return a.get(head);
    }
}

6 - 剑指 Offer 05. 替换空格

class Solution {
    public String replaceSpace(String s) {
        return s.replace(" ", "%20");
    }
}

7 - 剑指 Offer 58 - II. 左旋转字符串

class Solution {
    public String reverseLeftWords(String s, int n) {
        int m = s.length();
        StringBuilder a = new StringBuilder(m);
        int i, j = 0;
        for (i = n; j < m; j ++)
        {
            a.append(s.charAt(i));
            i = (i + 1) % m;
        }
        return a.toString();
    }
}

8 - 剑指 Offer 03. 数组中重复的数字

class Solution {
    public int findRepeatNumber(int[] nums) {
        int n = nums.length;
        boolean[] a = new boolean[n];
        for (int i = 0; i < n; i ++)
        {
            if (a[nums[i]] == false)
                a[nums[i]] = true;
            else
                return nums[i];
        }
        return -1;
    }
}

9 - 剑指 Offer 53 - I. 在排序数组中查找数字 I

class Solution {
    public int bSearch(int[] nums, int target)
    {
        int l = 0, n = nums.length;
        int r = n - 1, ans = -1;
        for (; l <= r; )
        {
            int mid = (l + r) / 2;
            if (target <= nums[mid])
            {
                ans = mid;
                r = mid - 1;
            }
            else
                l = mid + 1;
        }
        return ans;
    }
    public int search(int[] nums, int target) {
        return ((bSearch(nums, target + 1) == -1) ? nums.length : bSearch(nums, target + 1))
                - ((bSearch(nums, target) == -1) ? nums.length : bSearch(nums, target));
    }
}

10 - 剑指 Offer 53 - II. 0～n-1中缺失的数字

class Solution {
    public int missingNumber(int[] nums) {
        int l = 0, r = nums.length - 1, ans = -1;
        for (; l <= r; )
        {
            int mid = (l + r) / 2;
            if (nums[mid] == mid) l = mid + 1;
            else
            {
                ans = mid;
                r = mid - 1;
            }
        }
        return (ans == -1) ? nums.length : ans;
    }
}

11 - 剑指 Offer 04. 二维数组中的查找

class Solution {
    public boolean find(int[][] matrix, int l, int r, int u, int d, int target)
    {
        // System.out.println("" + l + "," + r + "," + u + "," + d);
        if (l < 0 || u < 0 || r >= matrix[0].length || d >= matrix.length || r < l || d < u)
            return false;
        if ((r - l + 1) * (d - u + 1) <= 4)
        {
            for (int i = l; i <= r; i ++)
                for (int j = u; j <= d; j ++)
                    if (matrix[j][i] == target)
                        return true;
            return false;
        }
        int mid_lr = (l + r) / 2;
        int mid_ud = (u + d) / 2;
        if (matrix[mid_ud][mid_lr] == target)
            return true;
        else if (matrix[mid_ud][mid_lr] < target)
            return find(matrix, mid_lr, r, mid_ud, d, target) ||
                find(matrix, l, mid_lr, mid_ud + 1, d, target) ||
                find(matrix, mid_lr, r, u, mid_ud - 1, target);
        else
            return find(matrix, l, mid_lr, u, mid_ud, target) ||
                find(matrix, mid_lr, r, u, mid_ud - 1, target) ||
                find(matrix, l, mid_lr, mid_ud + 1, d, target);
    }
    public boolean findNumberIn2DArray(int[][] matrix, int target) {
        if (matrix.length == 0)
            return false;
        int l = 0, u = 0, r = matrix[0].length - 1, d = matrix.length - 1;
        return find(matrix, l, r, u, d, target);
    }
}

12 - 剑指 Offer 11. 旋转数组的最小数字

class Solution {
    public int minArray(int[] numbers) {
        int n = numbers.length;
        int l = 0, r = n - 1, ans = numbers[0];
        for (; l <= r; )
        {
            int mid = (l + r) / 2;
            ans = Math.min(ans, numbers[mid]);
            if (numbers[mid] < numbers[r])
                r = mid - 1;
            else if (numbers[mid] > numbers[r])
                l = mid + 1;
            else
                r --;
        }
        return ans;
    }
}

13 - 剑指 Offer 50. 第一个只出现一次的字符

class Solution {
    public char firstUniqChar(String s) {
        int n = s.length();
        int[] cnt = new int[26];
        for (int i = 0; i < n; i ++)
            cnt[s.charAt(i) - 'a'] ++;
        for (int i = 0; i < n; i ++)
            if (cnt[s.charAt(i) - 'a'] == 1)
                return s.charAt(i);
        return ' ';
    }
}

14 - 面试题32 - I. 从上到下打印二叉树

class Solution {
    public int[] levelOrder(TreeNode root) {
        if (root == null)
            return new int[0];
        Deque<TreeNode> a = new ArrayDeque<>();
        ArrayList<Integer> b = new ArrayList<>();
        a.addLast(root);
        for (; !a.isEmpty(); )
        {
            TreeNode x = a.removeFirst();
            b.add(x.val);
            if (x.left != null) a.addLast(x.left);
            if (x.right != null) a.addLast(x.right);
        }
        int[] c = new int[b.size()];
        for (int i = 0; i < b.size(); i ++)
        {
            c[i] = b.get(i);
        }
        return c;
    }
}

15 - 剑指 Offer 32 - II. 从上到下打印二叉树 II

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> a = new ArrayList<>();
        List<TreeNode> b = new ArrayList<>();
        if (root == null)
            return a;
        b.add(root);
        for (; !b.isEmpty(); )
        {
            List<TreeNode> c = new ArrayList<>();
            List<Integer> d = new ArrayList<>();
            for (int i = 0; i < b.size(); i ++)
            {
                TreeNode x = b.get(i);
                d.add(x.val);
                if (x.left != null)
                    c.add(x.left);
                if (x.right != null)
                    c.add(x.right);
            }
            a.add(d);
            b = c;
        }
        return a;
    }
}

16 - 剑指 Offer 32 - III. 从上到下打印二叉树 III

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> a = new ArrayList<>();
        List<TreeNode> b = new ArrayList<>();
        int level = 0;
        if (root == null)
            return a;
        b.add(root);
        for (; !b.isEmpty(); level ++)
        {
            List<TreeNode> c = new ArrayList<>();
            List<Integer> d = new ArrayList<>();
            for (int i = 0; i < b.size(); i ++)
            {
                TreeNode x = b.get(i);
                if (x.left != null)
                    c.add(x.left);
                if (x.right != null)
                    c.add(x.right);
            }
            if (level % 2 == 1)
                for (int i = b.size() - 1; i >= 0; i --)
                    d.add(b.get(i).val);
            else
                for (int i = 0; i < b.size(); i ++)
                    d.add(b.get(i).val);
            a.add(d);
            b = c;
        }
        return a;
    }
}

17 - 剑指 Offer 26. 树的子结构

class Solution {
    public boolean S(TreeNode A, TreeNode B)
    {
        if (A == null && B != null)
            return false;
        if (B != null)
            return (A.val == B.val) && S(A.left, B.left) && S(A.right, B.right);
        else
            return true;
    }
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        return (A != null) && (B != null) && ((S(A, B) == true) || isSubStructure(A.left, B)
        || isSubStructure(A.right, B)); 
    }
}

18 - 剑指 Offer 27. 二叉树的镜像

class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if (root == null)
            return root;
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        mirrorTree(root.left);
        mirrorTree(root.right);
        return root;
    }
}

19 - 剑指 Offer 28. 对称的二叉树

class Solution {
    public boolean S(TreeNode A, TreeNode B)
    {
        if (A == null && B != null || A != null && B == null)
            return false;
        if (A == null && B == null)
            return true;
        return (A.val == B.val) && S(A.left, B.right) && S(A.right, B.left);
    }
    public boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        return S(root.left, root.right);
    }
}

20 - 剑指 Offer 10- I. 斐波那契数列

class Solution {
    public int fib(int n) {
        int[] F = new int[101];
        int mod = 1000000007;
        F[0] = 0; F[1] = 1;
        for (int i = 2; i <= n; i ++) F[i] = (F[i - 1] + F[i - 2]) % mod;
        return F[n];
    }
}

21 - 剑指 Offer 10- II. 青蛙跳台阶问题

class Solution {
    public int numWays(int n) {
        int[] F = new int[101];
        int mod = 1000000007;
        F[0] = 1; F[1] = 1;
        for (int i = 2; i <= n; i ++) F[i] = (F[i - 1] + F[i - 2]) % mod;
        return F[n];
    }
}

22 - 剑指 Offer 63. 股票的最大利润

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        if (n == 0) return 0;
        int[] F = new int[n + 1];
        F[n] = 0;
        int Ans = 0;
        for (int i = n - 1; i >= 0; i --)
            F[i] = Math.max(F[i + 1], prices[i]);
        for (int i = 0; i < n; i ++)
            Ans = Math.max(Ans, F[i + 1] - prices[i]);
        return Ans;
    }
}

23 - 剑指 Offer 42. 连续子数组的最大和

class Solution {
    public int maxSubArray(int[] nums) {
        int n = nums.length;
        int[] Dp1 = new int[n];
        int[] Dp2 = new int[n];
        Dp1[0] = Math.max(0, nums[0]);
        Dp2[0] = nums[0];
        for (int i = 1; i < n; i ++)
        {
            Dp1[i] = Math.max(nums[i], Dp1[i - 1] + nums[i]);
            Dp2[i] = Math.max(Dp2[i - 1], Dp1[i]);
        }
        return Dp2[n - 1];
    }
}

24 - 剑指 Offer 47. 礼物的最大价值

class Solution {
    public int maxValue(int[][] grid) {
        int n = grid.length;
        int m = grid[0].length;
        int[][] F = new int[n][m];
        F[0][0] = grid[0][0];
        for (int i = 0; i < n; i ++)
            for (int j = 0; j < m; j ++)
            {
                if (i != 0) F[i][j] = Math.max(F[i][j], F[i - 1][j] + grid[i][j]);
                if (j != 0) F[i][j] = Math.max(F[i][j], F[i][j - 1] + grid[i][j]);
            }
        return F[n - 1][m - 1];
    }
}

25 - 剑指 Offer 46. 把数字翻译成字符串

class Solution {
    public int translateNum(int num) {
        String s = String.valueOf(num);
        int n = s.length();
        int[] F = new int[n + 1];
        F[0] = 1;
        for (int i = 1; i <= n; i ++)
        {
            F[i] = F[i - 1];
            if (i >= 2 && Integer.valueOf(s.substring(i - 2, i)) >= 10
            && Integer.valueOf(s.substring(i - 2, i)) <= 25)
                F[i] += F[i - 2];
        }
        return F[n];
    }
}

26 - 剑指 Offer 48. 最长不含重复字符的子字符串

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        int ans = 0;
        int[] cnt = new int[256];
        int j = 0;
        for (int i = 0; i < n; i ++)
        {
            cnt[s.charAt(i)] ++;
            if (cnt[s.charAt(i)] == 2)
            {
                for (; j < i && s.charAt(i) != s.charAt(j); j ++)
                    cnt[s.charAt(j)] --;
                cnt[s.charAt(j ++)] --;
            }
            ans = Math.max(ans, i - j + 1);
            // System.out.println(i + " " + j);
        }
        return ans;
    }
}

27 - 剑指 Offer 18. 删除链表的节点

class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        ListNode i = head, j = new ListNode();
        for (; i != null; j = i, i = i.next)
            if (i.val == val)
                break;
        if (i == head)
            return head.next;
        else
        {
            j.next = i.next;
            return head;
        }
    }
}

28 - 剑指 Offer 22. 链表中倒数第k个节点

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode i = head, j = head;
        for (int s = 0; s < k; s ++, j = j.next);
        for (; i != null && j != null; i = i.next, j = j.next);
        return i;
    }
}

29 - 剑指 Offer 25. 合并两个排序的链表

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function(l1, l2) {
    var cur = new ListNode(-1);
    let head = cur;
    for (; l1 != null && l2 != null; )
    {
        if (l1.val < l2.val)
            cur.next = l1, l1 = l1.next;
        else
            cur.next = l2, l2 = l2.next;
        cur = cur.next;
    }
    for (; l1 != null; l1 = l1.next)
        cur.next = l1, cur = cur.next;
    for (; l2 != null; l2 = l2.next)
        cur.next = l2, cur = cur.next;
    return head.next;
};

30 - 剑指 Offer 52. 两个链表的第一个公共节点

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
    var S = new Set();
    for (; headA != null; headA = headA.next)
        S.add(headA);
    for (; headB != null; headB = headB.next)
        if (S.has(headB))
            return headB;
    return null;
};

31 - 剑指 Offer 21. 调整数组顺序使奇数位于偶数前面

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var exchange = function(nums) {
    var a = new Array();
    var l = 0;
    var r = nums.length - 1;
    for (let i = 0; i < nums.length; i ++)
    {
        if (nums[i] % 2 == 1)
            a[l ++] = nums[i];
        else
            a[r --] = nums[i];
    }
    return a;
};

32 - 剑指 Offer 57. 和为s的两个数字

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    var i = 0;
    var j = nums.length - 1;
    for (; i < j; )
    {
        if (nums[i] + nums[j] == target)
            return new Array(nums[i], nums[j]);
        else if (nums[i] + nums[j] < target)
            i ++;
        else
            j --;
    }
};

33 - 剑指 Offer 58 - I. 翻转单词顺序

/**
 * @param {string} s
 * @return {string}
 */
var reverseWords = function(s) {
    var a = s.split(" ").reverse();
    var S = new String();
    for (let i = 0; i < a.length; i ++)
    {
        if (a[i].length == 0) continue;
        S += " " + a[i];
    }
    return S.substring(1);
};

34 - 剑指 Offer 12. 矩阵中的路径

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
var exist = function(board, word) {
    var n = board.length;
    var m = board[0].length;
    var vis = new Array();
    for (let i = 0; i < n; i ++)
        vis[i] = new Array();
    function dfs(x, y, cur, fin)
    {
        if (vis[x][y] == true)
            return false;
        if (board[x][y] != word[cur])
            return false;
        if (cur == fin - 1)
            return true;
        vis[x][y] = true;
        var ans = false;
        if (y + 1 < m) ans |= dfs(x, y + 1, cur + 1, fin);
        if (!ans && x + 1 < n) ans |= dfs(x + 1, y, cur + 1, fin);
        if (!ans && x >= 1) ans |= dfs(x - 1, y, cur + 1, fin);
        if (!ans && y >= 1) ans |= dfs(x, y - 1, cur + 1, fin);
        vis[x][y] = false;
        return ans;
    }
    for (let i = 0; i < n; i ++)
        for (let j = 0; j < m; j ++)
            if (dfs(i, j, 0, word.length))
                return true;
    return false;
}

35 - 剑指 Offer 13. 机器人的运动范围

/**
 * @param {number} m
 * @param {number} n
 * @param {number} k
 * @return {number}
 */
var movingCount = function(m, n, k) {
    var dp = new Array(m);
    for (let i = 0; i < m; i ++)
        dp[i] = new Array(n);
    function cal(x)
    {
        let t = 0;
        for (; x; x = Math.floor(x / 10))
            t += x % 10;
        return t;
    }
    dp[0][0] = true;
    var ans = 0;
    for (let i = 0; i < m; i ++)
        for (let j = 0; j < n; j ++)
        {
            dp[i][j] = (cal(i) + cal(j) <= k) && ((i > 0 && dp[i - 1][j]) || (j > 0 && dp[i][j - 1]) || (i == 0 && j == 0));
            // console.log(i, j, dp[i][j], cal(i), cal(j), k);
            ans += (dp[i][j] == true) ? 1 : 0;
        }
    return ans;
};

36 - 剑指 Offer 34. 二叉树中和为某一值的路径

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} target
 * @return {number[][]}
 */
var pathSum = function(root, target) {
    var Ans = new Array();
    var cur = new Array();
    function dfs(x, y)
    {
        if (x == null)
            return;
        if (x.left == null && x.right == null)
        {
            if (y + x.val == target)
            {
                cur.push(x.val);
                Ans.push(cur.slice());
                cur.pop();
            }
            return;
        }
        cur.push(x.val);
        dfs(x.left, y + x.val);
        dfs(x.right, y + x.val);
        cur.pop();
    }
    dfs(root, 0);
    return Ans;
};

37 - 剑指 Offer 36. 二叉搜索树与双向链表

/**
 * // Definition for a Node.
 * function Node(val,left,right) {
 *    this.val = val;
 *    this.left = left;
 *    this.right = right;
 * };
 */
/**
 * @param {Node} root
 * @return {Node}
 */
var treeToDoublyList = function(root) {
    var s = null;
    var head = null;
    if (root == null)
        return null;
    function dfs(x)
    {
        if (x == null)
            return;
        dfs(x.left);
        if (head == null)
        {
            head = x;
            s = x;
        }
        else
        {
            x.left = s;
            s.right = x;
            s = x;
        }
        dfs(x.right);
    }
    dfs(root);
    // console.log(s, head == null);
    s.right = head;
    head.left = s;
    return head;
};

38 - 剑指 Offer 54. 二叉搜索树的第k大节点

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthLargest = function(root, k) {
    var Ans = null;
    function dfs(x)
    {
        if (x == null)
            return;
        dfs(x.right);
        k --;
        if (k == 0)
            Ans = x.val;
        dfs(x.left);
    }
    dfs(root);
    return Ans;
};

39 - 剑指 Offer 45. 把数组排成最小的数

/**
 * @param {number[]} nums
 * @return {string}
 */
var minNumber = function(nums) {
    var a = new Array();
    for (let i = 0; i < nums.length; i ++)
        a.push(nums[i].toString());
    a.sort(function(x, y){
        return x + y < y + x ? -1 : 1;
    });
    // console.log(a);
    return a.join("");
};

40 - 剑指 Offer 61. 扑克牌中的顺子

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var isStraight = function(nums) {
    // var x = (nums.map((x)=>(x > 0 ? x : 14)));
    // console.log(x);
    var a = new Array(14).fill(0);
    for (let i = 0; i < nums.length; i ++)
        a[nums[i]] ++;
    // var L = Math.min(...((nums.map((x)=>(x > 0 ? x : 14)))));
    // console.log(...nums, L);
    // var R = Math.max(...nums);
    // console.log(L, R);
    for (let i = 1; i <= 9; i ++)
    {
        let Flag = true;
        let b = a.slice();
        for (let j = i; j <= i + 4; j ++)
        {
            if (b[j] >= 1)
                b[j] --;
            else
                b[0] --;
            if (b[0] < 0)
            {
                Flag = false;
                break;
            }
        }
        if (Flag) return true;
    }
    return false;
};

41 - 剑指 Offer 40. 最小的k个数

/**
 * @param {number[]} arr
 * @param {number} k
 * @return {number[]}
 */
var getLeastNumbers = function(arr, k) {
    var a = new Array(10001).fill(0);
    var ans = new Array();
    for (let i = 0; i < arr.length; i ++)
        a[arr[i]] ++;
    for (let i = 0; i <= 10000; i ++)
        for (let j = 0; j < a[i] && k > 0; j ++)
        {
            ans.push(i);
            k --;
        }
    return ans;
};

42 - 剑指 Offer 41. 数据流中的中位数

/**
 * initialize your data structure here.
 */

class TreeNode {
    constructor(x)
    {
        this.val = x;
        this.left = null;
        this.right = null;
        this.size = 1;
    }
};

class Tree {
    ins(x, y)
    {
        if (x == null)
            return new TreeNode(y);
        x.size ++;
        if (y <= x.val)
            x.left = this.ins(x.left, y);
        else
            x.right = this.ins(x.right, y);
        // console.log("ins", x);
        if (Math.max(((x.left == null) ? 0 : x.left.size), ((x.right == null) ? 0 : x.right.size)) > 0.7 * x.size)
            return this.rebuild(x);
        else
            return x;
    }
    update(x)
    {
        x.size = ((x.left == null) ? 0 : x.left.size) + ((x.right == null) ? 0 : x.right.size) + 1;
    }
    build(l, r)
    {
        if (l > r) return;
        var mid = Math.floor((l + r) / 2);
        var x = new TreeNode();
        x.val = this.a[mid];
        x.left = this.build(l, mid - 1);
        x.right = this.build(mid + 1, r);
        this.update(x);
        return x;
    }
    dfs(x)
    {
        if (x == null)
            return;
        this.dfs(x.left);
        this.a.push(x.val);
        this.dfs(x.right);
    }
    rebuild(x)
    {
        this.a = new Array();
        this.dfs(x);
        return this.build(0, this.a.length - 1);
    }
    find(x, k)
    {
        // console.log("find", x, k);
        if (((x.left == null) ? 0 : x.left.size) + 1 == k)
            return x.val;
        else if (k <= ((x.left == null) ? 0 : x.left.size))
            return this.find(x.left, k);
        else
            return this.find(x.right, k - (((x.left == null) ? 0 : x.left.size) + 1));
    }
};

var root = null;
var T = new Tree();

var MedianFinder = function() {
    root = null;
    T = new Tree();
};

/** 
 * @param {number} num
 * @return {void}
 */
MedianFinder.prototype.addNum = function(num) {
    root = T.ins(root, num);
};

/**
 * @return {number}
 */
MedianFinder.prototype.findMedian = function() {
    var n = (root == null) ? 0 : root.size;
    // console.log("findMedian", n);
    if (n == 0)
        return null;
    if (n % 2 == 1)
        return T.find(root, Math.floor(n / 2) + 1);
    else
        return (T.find(root, n / 2) + T.find(root, n / 2 + 1)) / 2;
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * var obj = new MedianFinder()
 * obj.addNum(num)
 * var param_2 = obj.findMedian()
 */

43 - 剑指 Offer 55 - I. 二叉树的深度

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxDepth = function(root) {
    var ans = 0;
    function dfs(x, dep)
    {
        if (x == null)
        {
            ans = Math.max(ans, dep);
            return;
        }
        dfs(x.left, dep + 1);
        dfs(x.right, dep + 1);
    }
    dfs(root, 0);
    return ans;
};

44 - 剑指 Offer 55 - II. 平衡二叉树

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function(root) {
    var ans = true;
    function dfs(x)
    {
        if (ans == false)
            return;
        if (x == null)
            return 0;
        let left = dfs(x.left);
        let right = dfs(x.right);
        if (Math.abs(left - right) > 1) ans = false;
        return Math.max(left, right) + 1;
    }
    dfs(root);
    return ans;
};

45 - 剑指 Offer 64. 求1+2+…+n

/**
 * @param {number} n
 * @return {number}
 */
var sumNums = function(n) {
    function solve(n) {
        n && (n += solve(n - 1));
        return n;
    }
    return solve(n);
};

46 - 剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
    function solve(x, p, q)
    {
        if (x.val < p.val && x.val < q.val)
            return solve(x.right, p, q);
        else if (x.val > p.val && x.val > q.val)
            return solve(x.left, p, q);
        else
            return x;
    }
    return solve(root, p, q);
};

47 - 剑指 Offer 68 - II. 二叉树的最近公共祖先

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
    function solve(root, p, q)
    {
        if (root == null)
            return null;
        if (root == p || root == q)
            return root;
        var L = solve(root.left, p, q);
        var R = solve(root.right, p, q);
        if (L && R)
            return root;
        else return (L) ? L : R;
    }
    return solve(root, p, q);
};

48 - 剑指 Offer 07. 重建二叉树

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
var buildTree = function(preorder, inorder) {
    function solve(preO, inO)
    {
        if (preO.length == 0)
            return null;
        var x = preO[0];
        var node = new TreeNode(x);
        if (preO.length == 1)
            return node;
        for (var i = 0; i < inO.length; i ++)
            if (inO[i] == x)
                break;
        node.left = solve(preO.slice(1, i + 1), inO.slice(0, i));
        node.right = solve(preO.slice(i + 1), inO.slice(i + 1));
        return node;
    }
    return solve(preorder, inorder);
};

49 - 剑指 Offer 16. 数值的整数次方

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    var flag = false;
    if (n < 0)
    {
        flag = true;
        n = -n;
    }
    var ans = 1;
    for (; n; n = Math.floor(n / 2))
    {
        // console.log(n);
        if (n % 2 == 1) ans = ans * x;
        x = x * x;
    }
    return (flag ? 1. / ans : ans);
};

50 - 剑指 Offer 33. 二叉搜索树的后序遍历序列

/**
 * @param {number[]} postorder
 * @return {boolean}
 */
var verifyPostorder = function(postorder) {
    function solve(a)
    {
        if (a.length == 0)
            return true;
        var x = a[a.length - 1];
        for (var i = 0; i < a.length - 1; i ++)
            if (a[i] > x) break;
        var y = i;
        for (; i < a.length - 1; i ++)
            if (a[i] < x) return false;
        return solve(a.slice(0, y)) && solve(a.slice(y, a.length - 1));
    }
    return solve(postorder);
};

51 - 剑指 Offer 15. 二进制中1的个数

/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function(n) {
    var ans = 0;
    for (let i = 31; i >= 0; i --)
    {
        if (n & (1 << i))
            ans ++;
    }
    return ans;
};

52 - 剑指 Offer 65. 不用加减乘除做加法

/**
 * @param {number} a
 * @param {number} b
 * @return {number}
 */
var add = function(a, b) {
    var c = 0;
    var ans = 0;
    for (let i = 0; i < 32; i ++)
    {
        let x = a & (1 << i) ? 1 : 0;
        let y = b & (1 << i) ? 1 : 0;
        ans |= ((x ^ y ^ c) << i);
        c = (x & y) | (x & c) | (y & c);
    }
    return ans;
};

53 - 剑指 Offer 56 - I. 数组中数字出现的次数

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var singleNumbers = function(nums) {
    var ret = 0;
    for (var x of nums)
        ret ^= x;
    var i = 0;
    for (; !(ret & 1); ret >>= 1)
        i ++;
    var res1 = 0;
    var res2 = 0;
    for (var x of nums)
        if ((1 << i) & x)
            res1 ^= x;
        else
            res2 ^= x;
    return [res1, res2];
};

54 - 剑指 Offer 56 - II. 数组中数字出现的次数 II

/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function(nums) {
    var a = new Array(31).fill(0);
    for (var x of nums)
        for (let i = 0; i < 31; i ++)
            if (x & (1 << i))
                a[i] = (a[i] + 1) % 3;
    var ans = 0;
    for (let i = 0; i < 31; i ++)
        if (a[i])
            ans |= (1 << i);
    return ans;
};

55 - 剑指 Offer 39. 数组中出现次数超过一半的数字

/**
 * @param {number[]} nums
 * @return {number}
 */
var majorityElement = function(nums) {
    var num = 0;
    var numc = 0;
    for (let x of nums)
    {
        if (numc == 0)
            num = x;
        if (x == num)
            numc ++;
        else
            numc --;
    }
    return num;
};

56 - 剑指 Offer 66. 构建乘积数组

/**
 * @param {number[]} a
 * @return {number[]}
 */
var constructArr = function(a) {
    var b = new Array();
    var L = new Array();
    var R = new Array();
    if (a.length == 0)
        return [];
    if (a.length == 1)
        return [1];
    L[0] = a[0];
    for (let i = 1; i < a.length; i ++)
    {
        L[i] = L[i - 1] * a[i];
    }
    R[a.length - 1] = a[a.length - 1];
    for (let i = a.length - 2; i >= 0; i --)
    {
        R[i] = R[i + 1] * a[i];
    }
    b[0] = R[1];
    b[a.length - 1] = L[a.length - 2];
    for (let i = 1; i < a.length - 1; i ++)
    {
        b[i] = L[i - 1] * R[i + 1];
    }
    return b;
};

57 - 剑指 Offer 14- I. 剪绳子

/**
 * @param {number} n
 * @return {number}
 */
var cuttingRope = function(n) {
    // 2 * a + 3 * b
    // 2 * (a - 3) + 3 * (b + 2)
    // 3 * 3 / 2 / 2 / 2 > 1
    var ans = 1;
    if (n == 2)
        return 1;
    if (n == 3)
        return 2;
    for (; n >= 3; n -= 3)
        ans *= 3;
    if (n == 2) ans *= 2;
    if (n == 1) ans = ans / 3 * 4;
    return ans;
};

58 - 剑指 Offer 57 - II. 和为s的连续正数序列

/**
 * @param {number} target
 * @return {number[][]}
 */
var findContinuousSequence = function(target) {
    // i L i * L + L(L - 1) / 2
    var ans = [];
    for (let i = 1; 2 * i + 1 <= target; i ++)
    {
        let cur = 0;
        let a = new Array();
        for (let j = i; cur < target; j ++)
        {
            cur += j;
            a.push(j);
        }
        if (cur == target)
            ans.push(a);
    }
    return ans;
};

59 - 剑指 Offer 62. 圆圈中最后剩下的数字

/**
 * @param {number} n
 * @param {number} m
 * @return {number}
 */
var lastRemaining = function(n, m) {
    ans = 0;
    for (let i = 2; i <= n; i ++)
        ans = (ans + m) % i;
    return ans;
};

60 - 剑指 Offer 29. 顺时针打印矩阵

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function(matrix) {
    let n = matrix.length;
    if (n == 0)
        return [];
    let m = matrix[0].length;
    if (m == 0)
        return [];
    let cnt = 1;
    let dx = [0, 1, 0, -1];
    let dy = [1, 0, -1, 0];
    let A = [];
    let flag = new Array();
    for (let i = 0; i < n; i ++)
        flag[i] = new Array();
    let x = 0, y = -1;
    let c = 0;
    for (; cnt <= n * m; cnt ++)
    {
        nx = x + dx[c];
        ny = y + dy[c];
        while (nx < 0 || nx >= n || ny < 0 || ny >= m || flag[nx][ny] == true)
            c = (c + 1) % 4, nx = x + dx[c], ny = y + dy[c];
        x = nx;
        y = ny;
        flag[x][y] = true;
        A.push(matrix[x][y]);
    }
    return A;
};

61 - 剑指 Offer 31. 栈的压入、弹出序列

/**
 * @param {number[]} pushed
 * @param {number[]} popped
 * @return {boolean}
 */
var validateStackSequences = function(pushed, popped) {
    let A = [];
    let top = 0, i = 0;
    for (let x of pushed)
    {
        A[++ top] = x;
        while (top > 0 && A[top] == popped[i])
            top --, i ++;
    }
    if (top == 0)
        return true;
    else
        return false;
};

62 - 剑指 Offer 20. 表示数值的字符串

/**
 * @param {string} s
 * @return {boolean}
 */
var isNumber = function(s) {
    s = s.trim();
    function checkI(s)
    {
        if (s.charAt(0) == '+' || s.charAt(0) == '-')
            s = s.slice(1);
        if (s.length == 0)
            return false;
        for (let i = 0; i < s.length; i ++)
        {
            let c = s.charAt(i);
            if (c < '0' || c > '9') return false;
        }
        return true;
    }
    function checkS(s)
    {
        if (s.charAt(0) == '+' || s.charAt(0) == '-')
            s = s.slice(1);
        let dot = s.indexOf('.');
        if (s.length <= 1)
            return false;
        if (dot == -1)
            return false;
        // console.log(s, dot);
        for (let i = 0; i < dot; i ++)
        {
            let c = s.charAt(i);
            if (c < '0' || c > '9') return false;
        }
        for (let i = dot + 1; i < s.length; i ++)
        {
            let c = s.charAt(i);
            if (c < '0' || c > '9') return false;
        }
        return true;
    }
    let x = s.search("e|E");
    if (x == -1)
        return checkI(s) || checkS(s);
    else
        return (checkI(s.slice(0, x)) || checkS(s.slice(0, x))) &&
                (checkI(s.slice(x + 1)));
};

63 - 剑指 Offer 67. 把字符串转换成整数

/**
 * @param {string} str
 * @return {number}
 */
var strToInt = function(str) {
    str = str.trimLeft();
    if (str.length == 0 || str.trim().length == 0)
        return 0;
    let res = 0, F = 0;
    if (str.charAt(0) == '+') ;
    else if (str.charAt(0) == '-') F = 1;
    else if (str.charAt(0) >= '0' && str.charAt(0) <= '9');
    else return 0;
    // console.log(str);
    if (str.charAt(0) == '+' || str.charAt(0) == '-')
        str = str.slice(1);
    for (let i = 0; i < str.length; i ++)
    {
        let c = str.charAt(i);
        if (c < '0' || c > '9')
            return res * (F == 1 ? -1 : 1);
        res = res * 10 + (c - '0');
        res = Math.min(res, 2147483647 + F);
    }
    // console.log(res, F);
    return res * (F == 1 ? -1 : 1);
};

64 - 剑指 Offer 59 - I. 滑动窗口的最大值

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var maxSlidingWindow = function(nums, k) {
    let A = [], B = [];
    if (nums.length == 0)
        return [];
    for (let i = 0; i < k - 1; i ++)
    {
        for (; A.length >= 1 && A[A.length - 1].val <= nums[i]; A.pop());
        A.push({index : i, val: nums[i]});
    }
    for (let i = k - 1; i < nums.length; i ++)
    {
        for (; A.length >= 1 && A[A.length - 1].val <= nums[i]; A.pop());
        A.push({index : i, val: nums[i]});
        if (A[0].index <= i - k) A.shift();
        B.push(A[0].val);
    }
    return B;
};

65 - 剑指 Offer 59 - II. 队列的最大值

var MaxQueue = function() {
    this.a = [], this.b = [];
};

/**
 * @return {number}
 */
MaxQueue.prototype.max_value = function() {
    if (this.b.length == 0)
        return -1;
    else
        return this.b[0];
};

/** 
 * @param {number} value
 * @return {void}
 */
MaxQueue.prototype.push_back = function(value) {
    for (; this.b.length >= 1 && this.b[this.b.length - 1] < value; this.b.pop());
    this.b.push(value), this.a.push(value);
};

/**
 * @return {number}
 */
MaxQueue.prototype.pop_front = function() {
    if (this.a.length == 0)
        return -1;
    let c = this.a.shift();
    if (c == this.b[0]) this.b.shift();
    return c;
};

/**
 * Your MaxQueue object will be instantiated and called as such:
 * var obj = new MaxQueue()
 * var param_1 = obj.max_value()
 * obj.push_back(value)
 * var param_3 = obj.pop_front()
 */

66 - 剑指 Offer 37. 序列化二叉树

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

/**
 * Encodes a tree to a single string.
 *
 * @param {TreeNode} root
 * @return {string}
 */
var serialize = function(root) {
    let ans = new String();
    function dfs(x)
    {
        if (x == null)
        {
            ans = ans.concat(",null");
            return;
        }
        ans = ans.concat("," + x.val);
        dfs(x.left), dfs(x.right);
    }
    dfs(root);
    return ans.slice(1);
};

/**
 * Decodes your encoded data to tree.
 *
 * @param {string} data
 * @return {TreeNode}
 */
var deserialize = function(data) {
    let a = data.split(",");
    function dfs()
    {
        let c = a.shift();
        if (c == "null")
            return null;
        var x = new TreeNode(Number(c));
        x.left = dfs();
        x.right = dfs();
        return x;
    }
    var root = dfs();
    return root;
};

/**
 * Your functions will be called as such:
 * deserialize(serialize(root));
 */

67 - 剑指 Offer 38. 字符串的排列

/**
 * @param {string} s
 * @return {string[]}
 */
var permutation = function(s) {
    let a = Array.from(s).sort();
    let ans = [];
    let n = a.length;
    let next = true;
    for (; next; )
    {
        ans.push(a.join(""));
        for (var i = n - 1; i > 0; i --)
            if (a[i - 1] < a[i])
                break;
        i --;
        if (i < 0)
            next = false;
        for (var j = n - 1; j >= i; j --)
            if (a[j] > a[i])
                break;
        let t = a[i]; a[i] = a[j]; a[j] = t;
        let s1 = a.slice(0, i + 1);
        let s2 = a.slice(i + 1);
        a = s1.concat(s2.reverse());
    }
    return ans;
};

68 - 面试题19. 正则表达式匹配

class Solution {
    public boolean isMatch(String s, String p) {
        int n = s.length();
        int m = p.length();
        boolean[][] Dp = new boolean[n + 1][m + 1];
        Dp[0][0] = true;
        for (int i = 0; i <= n; i ++)
            for (int j = 0; j <= m; j ++)
            {
                if (j == 0) continue;
                if (p.charAt(j - 1) == '.')
                {
                    if (i >= 1 && j >= 1) Dp[i][j] |= Dp[i - 1][j - 1];
                }
                else if (p.charAt(j - 1) != '*')
                {
                    if (i >= 1 && j >= 1) Dp[i][j] |= Dp[i - 1][j - 1] && (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.');
                }
                else 
                {
                    if (j >= 2)
                        Dp[i][j] |= Dp[i][j - 2];
                    if (i >= 1 && j >= 2)
                        Dp[i][j] |= (Dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.'));
                }
                // System.out.println(i + " " + j + " " + Dp[i][j]);
            }
        return Dp[n][m];
    }
}

69 - 剑指 Offer 49. 丑数

/**
 * @param {number} n
 * @return {number}
 */
var nthUglyNumber = function(n) {
    let a = [1];
    let p1 = 0, p2 = 0, p3 = 0;
    for (; a.length < n; )
    {
        let x = a[p1] * 2, y = a[p2] * 3, z = a[p3] * 5;
        a.push(Math.min(x, y, z));
        if (x == a[a.length - 1]) p1 ++;
        if (y == a[a.length - 1]) p2 ++;
        if (z == a[a.length - 1]) p3 ++;
    }
    return a[n - 1];
};

70 - 剑指 Offer 60. n个骰子的点数

class Solution {
    public double[] dicesProbability(int n) {
        int[][] Dp = new int[n + 1][6 * n + 1];
        double[] answer = new double[6 * n - n + 1];
        int ans = 0;
        Dp[0][0] = 1;
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= 6; j ++)
                for (int k = i + j - 1; k <= 6 * i; k ++)
                    Dp[i][k] += Dp[i - 1][k - j];
        for (int i = n; i <= 6 * n; i ++)
            ans += Dp[n][i];
        for (int i = n; i <= 6 * n; i ++)
            answer[i - n] = 1.0 * Dp[n][i] / ans;
        return answer;
    }
}

71 - 剑指 Offer 17. 打印从1到最大的n位数

/**
 * @param {number} n
 * @return {number[]}
 */
var printNumbers = function(n) {
    let limit = 1;
    for (let i = 0; i < n; i ++) limit *= 10;
    a = new Array(limit - 1);
    for (let i = 1; i < limit; i ++)
        a[i - 1] = i;
    return a;
};

72 - 剑指 Offer 51. 数组中的逆序对

/**
 * @param {number[]} nums
 * @return {number}
 */
var reversePairs = function(nums) {
    var ans = 0;
    function Solve(l, r)
    {
        if (l >= r) return;
        let mid = Math.floor((l + r) / 2);
        Solve(l, mid), Solve(mid + 1, r);
        let L = l, R = mid + 1, cnt = 0, temp = [];
        for (; L <= mid && R <= r; )
        {
            if (nums[L] <= nums[R]) ans += R - 1 - mid, temp[cnt ++] = nums[L ++];
            else temp[cnt ++] = nums[R ++];
        }
        for (; L <= mid; L ++) temp[cnt ++] = nums[L], ans += R - 1 - mid;
        for (; R <= r; R ++) temp[cnt ++] = nums[R];
        for (let i = l; i <= r; i ++) nums[i] = temp[i - l];
        // console.log(l, r, nums, ans);
    }
    Solve(0, nums.length - 1);
    return ans;
};

73 - 剑指 Offer 14- II. 剪绳子 II

/**
 * @param {number} n
 * @return {number}
 */
var cuttingRope = function(n) {
    if (n == 2)
        return 1;
    else if (n == 3)
        return 2;
    else
    {
        let ans = 1, mod = 1000000007;
        for (; n > 4; n -= 3)
            ans = ans * 3 % mod;
        ans = ans * n % mod;
        return ans;
    }
};

74 - 剑指 Offer 43. 1～n 整数中 1 出现的次数

/**
 * @param {number} n
 * @return {number}
 */
var countDigitOne = function(n) {
    let res = 1, ans = 0;
    for (; res <= n; res *= 10)
    {
        let d = Math.floor(n / (res * 10));
        ans += d * res;
        let s = n % (res * 10);
        if (s >= 2 * res) ans += res;
        else if (s >= res) ans += s - res + 1;
    }
    return ans;
};

75 - 剑指 Offer 44. 数字序列中某一位的数字

/**
 * @param {number} n
 * @return {number}
 */
var findNthDigit = function(n) {
    let a = [1], s = 1;
    for (let i = 1; i <= 10; i ++)
        a[i] = (s * 10 - s) * i, s *= 10;
    for (let i = 1; i <= 10; i ++)
        a[i] += a[i - 1];
    for (var i = 10; i >= 0; i --)
        if (n >= a[i])
            break;
    let temp = Math.floor((n - a[i]) / (i + 1)), mod = (n - a[i]) % (i + 1);
    s = 1; for (let j = 0; j < i; j ++) s *= 10; s --;
    s += temp; s ++;
    // console.log(s);
    let b = [];
    for (; s; s = Math.floor(s / 10))
        b.unshift(s % 10);
    // console.log(b[mod]);
    return b[mod];
};