leetcode刷题全纪录（持续更新）
2.Add Two Numbers 原题链接https://leetcode.com/problems/add-two-numbers/
AC解：
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int sum = 0;
ListNode head = new ListNode(0);
ListNode dummy = new ListNode(0);
ListNode flag = dummy;
dummy.next = head;
while(l1 != null || l2 != null || sum >0) {
if(l1 != null) {
sum+= l1.val;
l1 = l1.next;
}
if(l2 != null) {
sum+= l2.val;
l2 = l2.next;
}
head.val = sum%10;
ListNode newNode = new ListNode(0);
head.next = newNode;
head = newNode;
flag = flag.next;
sum = sum/10;
}
if(flag.next.val==0) flag.next=null;
return dummy.next;
}
这个方法总会使链表尾出现一个多余的节点，优化后
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int sum = 0; ListNode dummy = new ListNode(0); ListNode flag = dummy; while(l1 != null || l2 != null || sum >0) { if(l1 != null) { sum+= l1.val; l1 = l1.next; } if(l2 != null) { sum+= l2.val; l2 = l2.next; } ListNode newNode = new ListNode(sum%10); flag.next = newNode; flag = flag.next; sum = sum/10; } return dummy.next; }
67. Add Binary
AC解：
public String addBinary(String a, String b) { int len1 = a.length()-1; int len2 = b.length()-1; int sum = 0; StringBuilder sb = new StringBuilder(); while(len1>=0||len2>=0||sum>0){ if(len1>=0){ int dA = a.charAt(len1)-'0'; sum += dA; len1--; } if(len2>=0){ int dB = b.charAt(len2)-'0'; sum += dB; len2--; } sb.append(sum%2); sum = sum/2; } return sb.reverse().toString(); }
7.Reverse Integer https://leetcode.com/problems/reverse-integer/
注意溢出
public int reverse(int x) {
int result = 0;
if(x ==0) return 0;
boolean flag = false;
if(x <0) {
x =Math.abs(x);
flag = true;
}
while(x >0){
//溢出处理
if(result!=0 && Integer.MAX_VALUE/result <10)
return 0;
result *=10;
result +=x%10;
x /=10;
}
if(flag)
result =-result;
return result;
}
8.Reverse bits
思路与反转十进制数类似
public int reverseBits(int n) {
int result = 0;
for (int i = 0; i < 32; i++) {
result += n & 1;
n >>>= 1; // CATCH: must do unsigned shift
if (i < 31) // CATCH: for last digit, don't shift!
result <<= 1;
}
return result;
}
338. Counting Bits
原题链接https://leetcode.com/problems/counting-bits/
自己的思路：Integer.bitCount()方法，但是原题并不推荐使用内嵌方法
bug free:遇到偶数时，其1的个数和该偶数除以2得到的数字的1的个数相同，遇到奇数时，其1的个数等于该奇数除以2得到的数字的1的个数再加1
public int[] countBits(int num) { int[] result = new int[num+1]; for (int i=1;i<num+1;i++) result[i]=result[i>>1]+(i&1); return result; }
413.Arithmetic Slices
题意：求一个数组中可以构成多少个等差数列
思路：参见http://blog.csdn.net/camellhf/article/details/52824234
public int numberOfArithmeticSlices(int[] A) { int count = 0; int incre = 0; for (int i = 2; i<A.length;i++) { if(A[i] - A[i-1] == A[i-1]-A[i-2]) count += ++incre; else incre=0; } return count; }
406. Queue Reconstruction by Height
题意，[h,k]分别表示该人的高度和他前面比他高的人数，将队列重排。
思路：先将输入数组排序，再插入排序；
public int[][] reconstructQueue(int[][] people) { if (people == null || people.length == 0 || people[0].length == 0) return new int[0][0]; //按照高度降序排列，高度相同时按照人数升序排列 Arrays.sort(people, new Comparator<int[]>() { public int compare(int[] a, int[] b) { if (b[0] == a[0]) return a[1] - b[1]; return b[0] - a[0]; } }); //插入排序思想，待插入的人总是低于（或等于）排好序的子队列的人，按照k的大小，将其插入到正确的位置 int n = people.length; ArrayList<int[]> tmp = new ArrayList<>(); for (int i = 0; i < n; i++) tmp.add(people[i][1], new int[]{people[i][0], people[i][1]}); int[][] res = new int[people.length][2]; int i = 0; for (int[] k : tmp) { res[i][0] = k[0]; res[i++][1] = k[1]; } return res; }
451.Sort Characters By Frequency 原题https://leetcode.com/problems/sort-characters-by-frequency/
思路：TopK---将字母及其个数存入hashmap中，利用优先队列PriorityQueue <Map.Entry<Character, Integer>>将这些键值对排序
public String frequencySort(String s) { Map<Character,Integer> map = new HashMap<> (); for (char c : s.toCharArray()) { if (map.containsKey(c)) { map.put(c, map.get(c) + 1); } else { map.put(c, 1); } } PriorityQueue <Map.Entry<Character, Integer>> pq = new PriorityQueue<>( new Comparator<Map.Entry<Character,Integer>>(){ @Override public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) { // TODO Auto-generated method stub return o2.getValue() - o1.getValue(); } } ); pq.addAll(map.entrySet()); StringBuilder sb = new StringBuilder(); while(!pq.isEmpty()) { Map.Entry<Character, Integer> m = pq.poll(); for(int i=0;i<m.getValue();i++){ sb.append(m.getKey()); } } return sb.toString();
347. Top K Frequent Element 原题链接 https://leetcode.com/problems/top-k-frequent-elements/
bug free:
public List<Integer> topKFrequent(int[] nums, int k) { List<Integer> list = new ArrayList<Integer>(); Map<Integer, Integer> map = new HashMap<>(); //map存储nums中元素和其个数 for (int c : nums) { if (map.containsKey(c)) { map.put(c, map.get(c) + 1); } else { map.put(c, 1); } } //优先队列按照自定义比较器对键值对进行排序 PriorityQueue<Map.Entry<Integer, Integer>> pq = new PriorityQueue<>( new Comparator<Map.Entry<Integer, Integer>>(){ @Override public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) { return o2.getValue() - o1.getValue(); } } ); pq.addAll(map.entrySet()); //根据参数k,出队K个元素 for (int i = 0; i < k; i++){ list.add(pq.poll().getKey()); } return list;
167. Two Sum II - Input array is sorted 原题链接 https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
public static int[] twoSum(int[] nums, int target) { int[] result = new int[2]; int i = 0,j = nums.length-1; int mid = target/2; while(i < j && nums[i]<=mid && nums[j]>=mid) { if(nums[i]+nums[j]==target) { result[0] = i+1; result[1] = j+1; break; } else if (nums[i] + nums[j] <target) i++; else if (nums[i] + nums[j] >target) j--; } return result; }
这个结果超过28%的solution 后来发现先j--再i++ 超过42%的solution~ 神奇
442. Find All Duplicates in an Array 原题链接：https://leetcode.com/problems/find-all-duplicates-in-an-array/
对于一个元素nums[i]，数组中nums[i]-1位置的元素反转为负数，遍历过程中如果遍历到负数，说明数字i+1出现了两次
public List<Integer> findDuplicates(int[] nums) { List<Integer> list = new ArrayList<Integer>(); for (int i = 0; i<nums.length; i++) { int index = Math.abs(nums[i])-1; if(nums[index]<0) list.add(Math.abs(index+1)); nums[index] = -nums[index]; } return list; }
86. Partition List 原题链接：https://leetcode.com/problems/partition-list/
思路：将链表根据val分为小于x和大于x两个子链表，再拼接
思路很简单，但是自己写的代码一直TLE，看了top solution里的答案才发现需要将big子链表末指针设为null
public ListNode partition(ListNode head, int x) { if(head == null) return head; ListNode small = new ListNode(0); ListNode big = new ListNode(0); ListNode smallHead = small; ListNode bigHead = big; //ListNode point = head; while(head != null) { if(head.val < x) { small.next = head; small = head; head = head.next; } else { big.next = head; big = head; head = head.next; } } big.next = null;//一定注意，之前没加这个语句一直TLE bigHead = bigHead.next; small.next = bigHead; return smallHead.next; }
Implement queue using stacks
public class MyQueue { private Stack<Integer> stack1 = new Stack<>();//辅助队尾入队，所有入队元素直接压入stack1 private Stack<Integer> stack2 = new Stack<>();//辅助队头出队 public void push(int x) { stack1.add(x); } public int pop(){ //如果stack2不为空，直接出栈 if(!stack2.empty()) return stack2.pop(); //如果stack2为空，将stack1中每个元素出栈后压入stack2,实现FIFO结构 while(!stack1.empty()) { stack2.push(stack1.pop()); } return stack2.pop(); } public int peek(){ if(!stack2.empty()) return stack2.peek(); while(!stack1.empty()) { stack2.push(stack1.pop()); } return stack2.peek(); } public boolean empty(){ return (stack1.empty() && stack2.empty()); } }
Implement stack using queues
public class MyStack { private Queue<Integer> q1 = new LinkedList<>(); private Queue<Integer> q2 = new LinkedList<>(); //push an element onto one stack public void push(int x) { if(!q1.isEmpty()) q1.add(x); else q2.add(x); } //pop from the top of the stack public int pop() { int result = 0; if(q1.isEmpty()) { while(q2.size() > 1) { q1.add(q2.poll()); } result = q2.poll(); } else{ while(q1.size() > 1) { q2.add(q1.poll()); } result = q1.poll(); } return result; } //get top element from stack public int top() { int result = 0; if(q1.isEmpty()) { while(q2.size() > 1) { q1.add(q2.poll()); } result = q2.poll(); q1.add(result); } else{ while(q1.size() > 1) { q2.add(q1.poll()); } result = q1.poll(); q2.add(result); } return result; } //find out whether the stack is empty public boolean empty() { return q1.isEmpty() && q2.isEmpty(); } }
使用一个队列实现栈
public class MyStack { private Queue<Integer> q = new LinkedList<Integer>(); // Push element x onto stack. public void push(int x) { q.add(x); for(int i = 1; i < q.size(); i ++) { //rotate the queue to make the tail be the head q.add(q.poll()); } } // Removes the element on top of the stack. public int pop() { return q.poll(); } // Get the top element. public int top() { return q.peek(); } // Return whether the stack is empty. public boolean empty() { return q.isEmpty(); } }
235. Lowest Common Ancestor of a Binary Search Tree
题意：求BST中两个节点的LCA，根据BST的特点递归求解
public TreeNode lowestCommonAncestor(TreeNode root,TreeNode p.TreeNode q) { if(root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left,p,q); if(root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p ,q); else return root; }
236.Lowest Common Ancestor of a Binary Tree
题意：求一个普通二叉树中两个节点的LCA
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root == null || root == p || root == q) return root; TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); if(left != null && right != null) return root; return left != null ? left : right; }
112.Path Sum
public class Solution { public boolean hasPathSum(TreeNode root, int sum) { if(root == null) return false; Stack<Integer> s = new Stack<Integer>(); int curSum = 0; return pathSum(root,sum,s,curSum); } private boolean pathSum(TreeNode root, int sum, Stack<Integer> s, int curSum) { // TODO 自动生成的方法存根 curSum += root.val; s.push(root.val); boolean isLeaf = (root.left == null && root.right == null); if(curSum == sum && isLeaf) { return true; } boolean left = false, right = false; if(root.left != null) left = pathSum(root.left,sum,s,curSum); if(root.right != null) right = pathSum(root.right,sum,s,curSum); s.pop(); return left || right; } }
113.Path Sum 2
找出二叉树中满足给定sum的路径和（从根节点到叶子），将结果放在list中
public class Solution { public List<List<Integer>> pathSum(TreeNode root, int sum) { List<List<Integer>> list = new LinkedList<List<Integer>>(); if(root == null) return list; Stack<Integer> s = new Stack<Integer>(); int curSum = 0; pathSum(root,list,sum,s,curSum); return list; } private void pathSum(TreeNode root, List<List<Integer>> list, int sum, Stack<Integer> s, int curSum) { // TODO 自动生成的方法存根 curSum += root.val; s.push(root.val); boolean isLeaf = (root.left == null && root.right == null); if(curSum == sum && isLeaf) { List<Integer> subList = new LinkedList<> (); for(Integer x : s ){ subList.add(x); } list.add(subList); } if(root.left != null) pathSum(root.left,list,sum,s,curSum); if(root.right != null) pathSum(root.right,list,sum,s,curSum); s.pop();//要返回到上一个节点 } }
437 path sum 3
public class Solution { public int pathSum(TreeNode root, int sum) { if(root == null) return 0; return findPath(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum); } public int findPath(TreeNode root, int sum){ int res = 0; if(root == null) return res; if(sum == root.val) res++; res += findPath(root.left, sum - root.val); res += findPath(root.right, sum - root.val); return res; } }
300. Longest Increasing Subsequence
数组中最长上升子序列的长度
https://leetcode.com/problems/longest-increasing-subsequence/discuss/
class Solution { public int lengthOfLIS(int[] nums) { int len = nums.length; if(len == 0) return 0; int[] dp = new int[len]; for(int i = 0 ; i < len; i++){ dp[i] = 1; } int max = 0; for(int i = 0; i < len; i++) { for(int j = 0; j < i; j++) { if(nums[i] > nums[j]){ dp[i] = Math.max(dp[i],dp[j]+1); } } max = Math.max(dp[i],max); } return max; } }
102. Binary Tree Level Order Traversal
二叉树层序遍历
public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<List<Integer>>(); Queue<TreeNode> queue = new LinkedList<>(); if(root == null) return res; queue.add(root); while(!queue.isEmpty()){ int levelNum = queue.size(); List<Integer> sub = new ArrayList<>(); for(int i = 0 ; i < levelNum; i++) { if(queue.peek().left != null) queue.offer(queue.peek().left); if(queue.peek().right != null ) queue.offer(queue.peek().right); sub.add(queue.poll().val); } res.add(sub); } return res; }
647.Palindromic Substrings
public int countSubstrings(String s) { int n = s.length(); int res = 0; boolean[][] dp = new boolean[n][n]; for (int i = n - 1; i >= 0; i--) { for (int j = i; j < n; j++) { //关键点：如果头尾相等，并且去掉头尾后也是回文 就是回文 特殊情况：j-i<3表示去掉头尾为空字符串 dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1]); if(dp[i][j]) ++res; } } return res; }
lintcode 92.背包问题
public class Solution { /** * @param m: An integer m denotes the size of a backpack * @param A: Given n items with size A[i] * @return: The maximum size */ public int backPack(int m, int[] A) { // write your code here int[][] dp = new int[A.length+1][m + 1]; for(int i = 0; i <= A.length; i++){ for(int j = 0; j <= m; j++){ if(i==0||j==0){ dp[i][j]=0; } else if(A[i-1] > j ){ dp[i][j] = dp[i-1][j]; } else { dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j-A[i-1]]+A[i-1]); } } } return dp[A.length][m]; } }
lintcode 125.背包问题2
public class Solution { /* * @param m: An integer m denotes the size of a backpack * @param A: Given n items with size A[i] * @param V: Given n items with value V[i] * @return: The maximum value */ public int backPackII(int m, int[] A, int[] V) { // write your code here int[][] dp = new int[A.length+1][m+1]; for(int i = 0; i <= A.length; i ++) { for(int j = 0; j <= m; j++) { if(i == 0 || j == 0){ dp[i][j] = 0; } else if(A[i-1] > j) { dp[i][j] = dp[i-1][j]; } else { dp[i][j] = Math.max(dp[i-1][j],dp[i-1][j-A[i-1]]+V[i-1]); } } } return dp[A.length][m]; } }
lintcode 97.最长公共子序列
public class Solution { /* * @param A: A string * @param B: A string * @return: The length of longest common subsequence of A and B */ public int longestCommonSubsequence(String A, String B) { // write your code here int m = A.length(); int n = B.length(); int[][] dp = new int[m+1][n+1]; int max = 0; for(int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { dp[i][j] =Math.max(dp[i-1][j],dp[i][j-1]); if(A.charAt(i-1) == B.charAt(j-1)) { dp[i][j] = dp[i-1][j-1]+1; } } } return dp[m][n]; } }
lintcode 79.最长公共子串
public class Solution { /* * @param A: A string * @param B: A string * @return: the length of the longest common substring. */ public int longestCommonSubstring(String A, String B) { // write your code here int m = A.length(); int n = B.length(); int[][] dp = new int[m+1][n+1]; //f[i][j] int max = 0; for (int i = 1; i <= m; i++){ for (int j = 1; j <= n; j++) { if(A.charAt(i-1) == B.charAt(j-1)) dp[i][j] = dp[i-1][j-1] + 1; else dp[i][j] = 0; max = Math.max(dp[i][j], max); } } return max; } }
lintcode 110.最小路径和
public class Solution { /* * @param grid: a list of lists of integers * @return: An integer, minimizes the sum of all numbers along its path */ public int minPathSum(int[][] grid) { // write your code here for(int i = 0; i < grid.length;i++){ for(int j = 0; j < grid[0].length; j++){ if(i == 0 && j == 0){} else if (i == 0 && j > 0) { grid[i][j] += grid[i][j-1]; } else if(i > 0 && j == 0){ grid[i][j] += grid[i-1][j]; } else { grid[i][j] += Math.min(grid[i][j-1],grid[i-1][j]); } } } return grid[grid.length-1][grid[0].length-1]; } }
