每日一题

1 leetCode 273 Integer to English Words

思路三个数一起，分3个特殊数组

 1 public class Solution {
 2     private final String[] LESS_THAN_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve",
 3                                            "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
 4     private final String[] TENS = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
 5     private final String[] THOUSANDS = {"", "Thousand", "Million", "Billion"};
 6     public String numberToWords(int num) {
 7         if (num == 0) return "Zero";
 8         int i = 0;
 9         String word = "";
10         while (num > 0){
11             if (num % 1000 != 0){
12                 word = help(num % 1000) + THOUSANDS[i] + " " + word;
13             }
14             i++;
15             num /= 1000;
16         }
17         return word.trim();
18     }
19     
20     
21     private String help(int n) {
22         if (n == 0) return "";
23         else if (n < 20){
24             return LESS_THAN_20[n] + " "; 
25         } else if (n < 100) {
26             return TENS[n / 10] + " " + help(n % 10);
27         } else {
28             return LESS_THAN_20[n / 100] + " Hundred " + help(n % 100);
29         }
30     }
31 }

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2 592 Fraction Addition and Subtration

每次读入3个字母

public class Solution {
    public String fractionAddition(String expression) {
         Scanner sc = new Scanner(expression).useDelimiter("/|(?=[-+])");
        int A = 0, B = 1;
        while (sc.hasNext()){
            int a = sc.nextInt(), b = sc.nextInt();
            A = A * b + B * a;
            B = B * b;
            int g = gcd(A, B);
            A /= g;
            B /= g;
        }
        return A + "/" + B;
    }
    public int gcd(int a, int b){
        return a == 0 ? Math.abs(b) : gcd(b % a, a);
    }
}

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3 275 H-Index|| d

思路，二分查找，找到a[m] = len - m;

public class Solution {
    public int hIndex(int[] cit) {
        int len = cit.length;
        int l = 0, r = len - 1;
        while (l <= r) {
            int m = (l + r) / 2;
            if (cit[m] == len - m) return cit[m];
            else if (cit[m] > len - m) r = m - 1;
            else l = m + 1;
        }
        return len - l;
    }
}

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4 278 First Bad Version

二分，<

    public int firstBadVersion(int n) {
        int l = 1, r = n;
        while (l < r) {
            int m = (r - l) / 2 + l;
            if (isBadVersion(m)) {
                r = m;
            } else {
                l = m + 1; 
            } 
        }
        return r;
    }

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5 279 Perfect Squares

1 递归超时

    public int numSquares(int n) {
        int res = n, num = 2;
        while (num * num <= n) {
            int a = n / (num * num), b = n % (num * num);
            res = Math.min(res, a + numSquares(b));
        }
        return res;
    }

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2 数学

    public int numSquares(int n) {
        if(n % 4 == 0) n/= 4;
        if (n % 8 == 7) return 4;
        for (int x = 0; x * x <= n; x++){
            int y = (int)Math.sqrt(n - x * x);
            if (x * x + y * y == n){
                int res = 0;
                if (x > 0) res++;
                if (y > 0) res++;
                return res;
            }
        }
        return 3;

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3 动态规划

    public int numSquares(int n) {
        int[] res = new int[n + 1];
        Arrays.fill(res, 5);
        res[0] = 0;
        for (int i = 0; i <= n; i++) {
            for (int j = 1; j * j + i <= n; j++) {
                res[j * j + i] = Math.min(res[j * j + i], res[i] + 1);
            }
        }
        return res[n];
    }

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6 282 Expression And Operator

递归

public class Solution {
    List<String> res = null;
    public List<String> addOperators(String num, int target) {
        res = new ArrayList<>();
        if (num == null || num.length() == 0) return res;
        help(num, target, "", 0, 0, 0);
        return res;
    }
    public void help(String num, int target, String path, int pos, long val, long mul){
        if (pos == num.length() && target == val) {
            res.add(path);
            return;
        }
        for (int i = pos; i < num.length(); i++) {
            if (i != pos && num.charAt(pos) == '0') break;
            long cur = Long.parseLong(num.substring(pos, i + 1));
            if (pos == 0){
                help(num, target, path + cur, i + 1, cur, cur);
            } else{
                help(num, target, path + "+" + cur, i + 1, val + cur, cur);
                help(num, target, path + "-" + cur, i + 1, val - cur, -cur);
                help(num, target, path + "*" + cur, i + 1, val - mul + mul * cur, cur *mul);
            }
        }
    }
}

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7 283 Move Zeros

把不是0的移到前面

public class Solution {
    public void moveZeroes(int[] nums) {
        if (nums == null || nums.length == 0) return;
        int index = 0;
        for (int i : nums){
            if (i != 0) {
                nums[index++] = i;
            }
        }
        for (; index < nums.length; index++){
            nums[index] = 0;
        }
    }
}

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8 284 Peeking Iterator

next 变量

class PeekingIterator implements Iterator<Integer> {

    Iterator<Integer> itor;
    Integer next = null;
    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        itor = iterator;
        if (itor.hasNext()){
            next = itor.next();
        }
    }

    // Returns the next element in the iteration without advancing the iterator.
    public Integer peek() {
        return next;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    @Override
    public Integer next() {
        Integer item = next;
        if (itor.hasNext()){
            next = itor.next();
        } else {
            next = null;
        }
        return item;
    }

    @Override
    public boolean hasNext() {
        return next != null;
    }
}

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9 289 Game of Life

4 个状态

    public void gameOfLife(int[][] board) {
        int m = board.length, n = board[0].length;
        int[] dx = {-1, -1, -1, 0, 1, 1,  1, 0};
        int[] dy = {-1,  0,  1, 1, 1, 0, -1, -1};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int cou = 0;
                for (int k = 0; k < dx.length; k++){
                    int x = i + dx[k], y = j + dy[k];
                    if (x < 0 || x >= m || y < 0 || y >= n) continue;
                    if (board[x][y] == 1 || board[x][y] == 2) cou++;
                }
                if (board[i][j] == 1 && (cou < 2 || cou > 3)) board[i][j] = 2;
                else if (board[i][j] == 0 && cou == 3) board[i][j] = 3;
            }
        }
        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++) {
                board[i][j] %= 2;
            }
        }
    }

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10 word Pattern

hashmap 如果有检测是否相同，如果没检测值

一一映射

    public boolean wordPattern(String pattern, String str) {
        String[] arr = str.split(" ");
        HashMap<Character, String> map = new HashMap<>();
        if (pattern.length() != arr.length) return false;
        for (int i = 0; i < arr.length; i++) {
            char c = pattern.charAt(i);
            if (map.containsKey(c)){
                if (!map.get(c).equals(arr[i])){
                    return false;
                }
            } else {
                if (map.containsValue(arr[i])){
                    return false;
                }
                map.put(c, arr[i]);
            }
        }
        return true;
    }

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11 292 Nim Game

余4

   public boolean canWinNim(int n) {
        return n % 4 != 0;
    }

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12 295 Find Mediam from Data Stream

两个堆默认最小堆

public class MedianFinder {

    /** initialize your data structure here. */
    private PriorityQueue<Long> small = new PriorityQueue<>();
    private PriorityQueue<Long> large = new PriorityQueue<>();
    public MedianFinder() {
        
    }
    
    public void addNum(int num) {
        large.add((long)num);
        small.add(-large.poll());
        if (large.size() < small.size()){
            large.add(-small.poll());
        }
    }
    
    public double findMedian() {
        return large.size() > small.size() ? large.peek() : (large.peek() - small.peek()) / 2;
    }
}

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13 297 Serialize and Deseralize Binary Tree

前中后层遍历了

public class Codec {

    private static final String split = ",";
    private static final String NN = "X";
    
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        ser(root, sb);
        return sb.toString();
    }
    public void ser(TreeNode root, StringBuilder sb) {
        if (root == null) {
            sb.append(NN).append(split);
            return;
        }
        sb.append(root.val).append(split);
        ser(root.left, sb);
        ser(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        Deque<String> nodes = new LinkedList<>();
        nodes.addAll(Arrays.asList(data.split(split)));
        return des(nodes);
    }
    public TreeNode des(Deque<String> nodes){
        String val = nodes.remove();
        if (val.equals(NN)) return null;
        TreeNode root = new TreeNode(Integer.valueOf(val));
        root.left = des(nodes);
        root.right = des(nodes);
        return root;
    }
}

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public class Codec {

    
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder res = new StringBuilder();
        Queue<TreeNode> q = new LinkedList<>();
        if (root == null) return "";
        q.offer(root);
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            if (node == null) {
                res.append("n ");
                continue;
            }
            res.append(node.val + " ");
            q.offer(node.left);
            q.offer(node.right);
        }
        return res.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == "") return null; 
        Deque<TreeNode> nodes = new LinkedList<>();
        String[] values = data.split(" ");
        TreeNode root = new TreeNode(Integer.parseInt(values[0]));
        nodes.add(root);
        for (int i = 1; i < values.length; i++) {
            TreeNode node = nodes.poll();
            if (!values[i].equals("n")){
                TreeNode left = new TreeNode(Integer.parseInt(values[i]));
                node.left = left;
                nodes.add(left);
            }
            if (!values[++i].equals("n")){
                TreeNode right = new TreeNode(Integer.parseInt(values[i]));
                node.right = right;
                nodes.add(right);
            }
        }
        return root;
    }
}

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14 299 Bulls and Cows

    public String getHint(String secret, String guess) {
        int[] num = new int[10];
        int a = 0, b = 0;
        for (int i = 0; i < secret.length(); i++) {
            char c = secret.charAt(i);
            if (c == guess.charAt(i)) {
                a++;
            }
            num[c - '0']++;
        }
        for (int i = 0; i < guess.length(); i++) {
            if (num[guess.charAt(i) - '0'] > 0) {
                num[guess.charAt(i) - '0']--;
                b++;
            }
        }
        b -= a;
        return a + "A" + b + "B";
    }

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15 Longest Increasing Subsequence

动态规划dp[i] = Math.max(dp[i], dp[j] + 1)

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums.length == 0) return 0;
        int[] dp = new int[nums.length];
        int res = 0;
        Arrays.fill(dp, 1);
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]){
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            res = Math.max(res, dp[i]);
        }
        return res;
    }
}

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如果遍历到的新元素比ends数组首元素大，比尾元素小时，此时用二分查找法找到第一个不小于此新元素的位置，覆盖掉位置的原来的数字，以此类推直至遍历完整个nums数组，此时ends数组的长度就是我们要求的LIS的长度，特别注意的是ends数组的值可能不是一个真实的LIS，

    public int lengthOfLIS(int[] nums) {
        if (nums.length == 0) return 0;
        int[] num = new int[nums.length];
        num[0] = nums[0];
        int i = 0;
        for (int n : nums){
            if (num[i] < n) num[++i] = n;
            else if (num[0] > n) num[0] = n;
            else {
                int l = 0, r = i;
                while (l < r) {
                    int m = (r - l) / 2 + l;
                    if (num[m] < n) l = m + 1;
                    else r = m;
                }
                num[r] = n;
            }
        }
        return i + 1;
    }

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16 301 Remove Invalid Parenthese

bfs 多分树

    public List<String> removeInvalidParentheses(String s) {
        List<String> res = new ArrayList<>();
        Set<String> set = new HashSet<>();
        Queue<String> q = new LinkedList<>();
        if (s == null) return res;
        q.offer(s);
        set.add(s);
        boolean V = false;
        while (!q.isEmpty()){
            String str = q.poll();
            if (isV(str)){
                res.add(str);
                V = true;
            }
            if (V) continue;
            for (int i = 0; i < str.length(); i++) {
                if (str.charAt(i) != '(' && str.charAt(i) != ')') continue;
                String t = str.substring(0, i) + str.substring(i + 1, str.length());
                if (set.add(t)) {
                    q.offer(t);
                }
            }
        }
        return res;
    }
    public boolean isV(String s) {
        int count = 0;
        for (char c : s.toCharArray()){
            if (c == '(') count++;
            else if (c == ')' &&  count-- == 0) return false;
        }
        return count == 0;
    }

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    public List<String> removeInvalidParentheses(String s) {
        List<String> res = new ArrayList<>();
        help(s, res, 0, 0, new char[]{'(', ')'});
        return res;
    }
    public void help(String s, List<String> res, int last_i, int last_j, char[] par) {
        for (int count = 0, i = last_i; i < s.length(); i++) {
            if (s.charAt(i) == par[0]) count++;
            if (s.charAt(i) == par[1]) count--;
            if (count >= 0) continue;
            for (int j = last_j; j <= i; j++){
                if (s.charAt(j) == par[1] && (j == last_j || s.charAt(j - 1) != par[1])){
                    help(s.substring(0, j) + s.substring(j + 1), res, i, j, par);
                }
            }
            return;
        }
        String reverse = new StringBuilder(s).reverse().toString();
        if (par[0] == '('){
            help(reverse, res, 0, 0, new char[]{')', '('});
        } else {
            res.add(reverse);
        }
    }
}

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17303 Range sum Query

sum[j] - sum[i - 1];

public class NumArray {

    int[] nums;
    public NumArray(int[] nums) {
        this.nums = nums;
        for (int i = 1; i < nums.length; i++) {
            nums[i] = nums[i - 1] + nums[i];
        }
    }
    
    public int sumRange(int i, int j) {
        if (i == 0) {
            return nums[j];
        } else {
            return nums[j] - nums[i - 1];
        }
    }
}

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18 304 Range sum Query 2D

上一题

public class NumMatrix {

    int[][] dp;
    public NumMatrix(int[][] mat) {
        if (mat == null 
           || mat.length == 0 
           || mat[0].length == 0) return;
        int m = mat.length, n = mat[0].length;
        dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j - 1] - dp[i-1][j-1] + mat[i - 1][j - 1];
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        return dp[row2 + 1][col2 + 1] - dp[row1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1];
    }
}

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19 306 Additive Number

确定前两个数，剩下递推

    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        for (int i = 1; i <= n / 2; i++) {
            for (int j = 1; Math.max(i, j) <= n - i - j; j++) {
                if (isV(i, j, num)) return true;
            }
        }
        return false;
    }
    public boolean isV(int i, int j, String num) {
        if (i > 1 && num.charAt(0) == '0') return false;
        if (j > 1 && num.charAt(i) == '0') return false;
        String sum;
        Long x1 = Long.parseLong(num.substring(0, i));
        Long x2 = Long.parseLong(num.substring(i, j + i));
        for (int start = i + j; start != num.length(); start += sum.length()){
            x2 = x1 + x2;
            x1 = x2 - x1;
            sum = x2.toString();
            if (!num.startsWith(sum, start)) return false;
        }
        return true;
    }

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20 309 Best Time to Buy and Sell Stock with Cooldown

动态规划两状态持有，未持有

    public int maxProfit(int[] prices) {
        if (prices.length == 0) return 0;
        int b0 = - prices[0], b1 = b0;
        int s0 = 0, s1 = 0, s2 = 0;
        for (int i = 1; i < prices.length; i++) {
            b0 = Math.max(b1, s2 - prices[i]);
            s0 = Math.max(s1, b1 + prices[i]);
            b1 = b0; s2 = s1; s1 = s0;
        }
        return s0;
    }

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21 310 Minimum Heiht Trees

剥洋葱法

    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        if (n == 1) return Collections.singletonList(0);
        List<Integer> res = new ArrayList<>();
        List<Set<Integer>> graf = new ArrayList<>();
        for (int i = 0; i < n; i++) graf.add(new HashSet());
        for (int i = 0; i < edges.length; i++) {
            graf.get(edges[i][0]).add(edges[i][1]);
            graf.get(edges[i][1]).add(edges[i][0]);
        }
        for (int i = 0; i < graf.size(); i++) {
            if (graf.get(i).size() == 1) res.add(i);
        }
        while (n > 2){
            n -= res.size();
            List<Integer> cur = new ArrayList<>();
            for (int i : res){
                int t = graf.get(i).iterator().next();
                graf.get(t).remove(i);
                if (graf.get(t).size() == 1) {
                    cur.add(t);
                }
            }
            res = cur;
        }
        return res;
    }

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22 312 Burst Ballons 动态规划，不含边界

class Solution {
    public int maxCoins(int[] nums) {
        int[] num = new int[nums.length + 2];
        num[0] = num[nums.length + 1] = 1;
        for (int i = 1; i <= nums.length; i++) {
            num[i] = nums[i - 1];
        }
        int n = num.length;
        int[][] dp = new int[n][n];
        for (int i = 2; i < n; i++) {
            for (int left = 0; left < n - i; left++) {
                int right = left + i;
                for (int k = left + 1; k < right; k++) {
                    dp[left][right] = Math.max(dp[left][right], num[left] * num[k] * num[right] + dp[left][k] + dp[k][right]);
                }
            }
        }
        return dp[0][n - 1];
    }
}

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23 313 Super Ugly Number

   public int nthSuperUglyNumber(int n, int[] primes) {
        if (primes.length == 0) return 0;
        int[] res = new int[n], num = new int[primes.length];
        res[0] = 1;
        for (int i = 1; i < n; i++) {
            res[i] = Integer.MAX_VALUE;
            for (int j = 0; j < primes.length; j++) {
                res[i] = Math.min(res[i], res[num[j]] * primes[j]);
            }
            for (int j = 0; j < primes.length; j++) {
                if (res[i] == res[num[j]] * primes[j]) num[j]++;
            }
        }
        return res[n - 1];
    }

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24 315 Count of Smaller Numbers After Self

二叉搜索树，比它小的数，重复数

class Solution {
    class Node {
        int val, sum, dup;
        Node(int v, int s) {
            val = v;
            sum = s;
        }
    }
    public List<Integer> countSmaller(int[] nums) {
        int[] res = new int[nums.length];
        Node root = null;
        for (int i = nums.length - 1; i >= 0; i--) {
            root = insert(nums[i], res, i, root, 0);
        }
        return Arrays.asList(res);
    }
    public Node insert(int num, int[] res, int i, Node node, int presum) {
        if (node == null) {
            node = new Node(num, presum);
            res[i] = presum;
        } else if (node.val == num) {
            node.dup++;
            res[i] = presum + node.sum;
        } else if (node.val < num) {
            node.right = insert(num, res, i, node.right, presum + node.presum + dup);
        } else{
            node.sum++;
            node.left = insert(num, res, i, node.left, presum);
        }
        return node;
    }
}

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25 671. Second Minimum Node In a Binary Tree

    public int findSecondMinimumValue(TreeNode root) {
        if (root == null) {
            return -1;
        }
        int res = find(root, root.val);
        return res == Integer.MAX_VALUE ? -1 : res;
    }
    int find(TreeNode root, int val) {
        if (root == null) {
            return Integer.MAX_VALUE;
        }
        if (root.val != val) {
            return root.val;
        }
        return Math.min(find(root.left, val), find(root.right, val));
    }

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26 669. Trim a Binary Search Tree

class Solution {
    public TreeNode trimBST(TreeNode root, int L, int R) {
        if (root == null) {
            return root;
        }
        if (root.val > R) {
            return trimBST(root.left, L, R);
        }
        if (root.val < L) {
            return trimBST(root.right, L, R);
        }
        root.left = trimBST(root.left, L, R);
        root.right = trimBST(root.right, L, R);
        return root;
    }

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27 n

posted on 2017-08-03 09:42 wheleetcode 阅读(160) 评论(0) 编辑收藏举报