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Letcode Top 100链表专题

letcode top 100

Letcode动态规划专题-困难

DFS BFS 回溯 递归

LetCoce Top 100

class Solution { public int numIslands(char[][] grid) { if(grid == null || grid.length == 0) { return 0; } if (grid[0] == null || grid[0].length == 0) { return 0; } int row = grid.length;.

/*// Definition for a Node.class Node { public int val; public List<Node> neighbors; public Node() { val = 0; neighbors = new ArrayList<Node>(); } public Node(int _val) { val = _val; neighb.

方法一：hashSetclass Solution { public int lengthOfLongestSubstring(String s) { if(s==null||s.length()==0){ return 0; } Set<Character> set=new HashSet<>(); int max=0; int current=0; .

方法一：暴力破解的方法class Solution { public int subarraySum(int[] nums, int k) { if(nums==null||nums.length==0){ return 0; } int sum=0; int count=0; for(int i=0;i<nums.length;i++){ for(int j=.

class Solution { public boolean wordBreak(String s, List<String> wordDict) { Set<String> wordDictSet=new HashSet<>(wordDict); //定义一个集合 boolean[] dp=new boolean[s.length()+1]; //定义一个动态规划数组 dp[0]=tr.

class Solution { public int longestCommonSubsequence(String text1, String text2) { int m=text1.length(); int n=text2.length(); int[][] dp=new int[m+1][n+1]; for(int i=1;i<=text1.length();i++){ for(int j=.

方法一：升级版本 动态规划class Solution { public int uniquePaths(int m, int n) { int[][] dp=new int[m][n]; for(int i=0;i<m;i++){ dp[i][0]=1; } for(int j=0;j<n;j++){ dp[0][j]=1; } f.

方法一：动态规划+数组class Solution { public int rob(int[] nums) { if(nums==null||nums.length==0){ return 0; } int[] dp=new int[nums.length+1]; dp[0]=0; dp[1]=nums[0]; for(int i=2;i<=nums.length.

class Solution { public int fib(int n) { int[] arr=new int[n+1]; if(n==0||n==1){ return n; } if(n==2){ return 1; } arr[0]=0; arr[1]=1; arr[2]=1; for(int i=.

本题目一共有三种思路方法一：递归方法二：迭代class Solution { public int climbStairs(int n) { int[] arr=new int[n+1]; if(n==1||n==2){ return n; } arr[1]=1; arr[2]=2; for(int i=3;i<=n;i++){ arr[i].

class Solution { public String intToRoman(int num) { String[] I={"","I","II","III","IV","V","VI","VII","VIII","IX"}; String[] X={"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"}; String[] C={"","C","CC","CCC","CD","D","DC","D.

class Solution { public String reverseWords(String s) { if(s==null){ return null; } StringBuilder sb=new StringBuilder(); String[] ss=s.split(" "); for(int i=ss.length-1;i>=0;i--){ if(.

class Solution { public String reverseWords(String s) { if(s==null){ return s; } StringBuilder sb=new StringBuilder(); String[] ss=s.split(" "); for(int i=0;i<ss.length;i++){ if(ss[i].

使用最大堆class Solution { public int findKthLargest(int[] nums, int k) { if(nums.length==0||nums==null||k>nums.length){ return 0; } PriorityQueue<Integer> queue=new PriorityQueue<>((n1,n2)->(n2-n1)).

class MedianFinder { /** initialize your data structure here. */ PriorityQueue<Integer> maxHeap; //大顶堆 PriorityQueue<Integer> minHeap; //小顶堆 public MedianFinder() { maxHeap=new PriorityQueue<>((n1,n2)->(n2-n1.

思考边界条件/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */class Solution { public TreeNode lowestCommonAncestor(TreeNode root, .

递归的方式/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode righ.

做题思路：层序遍历/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode .

/** * 统一一下 * @param root * @return */ //前序 public static List<Integer> preOrder(TreeNode root){ List<Integer> list = new ArrayList(); Stack<TreeNode> stack = new Stack(); TreeNode cu.

方法一：递归的方法/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode .

方法一：递归class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> arraylist=new ArrayList<Integer>(); if(root==null){ return arraylist; } List<Integer> le.

方法一：递归方式class Solution { public List<Integer> preorderTraversal(TreeNode root) { List<Integer> result=new ArrayList<Integer>(); if(root==null){ return result; } List<Integer> left=.

方法一：顺序遍历class Solution { public int trap(int[] height) { if(height==null||height.length<3){ return 0; } int count=0; for(int i=1;i<height.length-1;i++){ int maxLeft=0; .

方法一：暴力破解的方式class Solution { public int triangleNumber(int[] nums) { if(nums.length==0||nums.length<3){ return 0; } int count=0; Arrays.sort(nums); for(int i=0;i<nums.length-2;i++){ .

方法一：暴力破解的方式class Solution { public List<List<Integer>> threeSum(int[] nums) { Set<List<Integer>> result=new HashSet<>(); if(nums.length==0||nums==null){ return new ArrayList<List<Integer.

//双指针class Solution { public int[] twoSum(int[] numbers, int target) { int[] arr=new int[2]; //权限校验 if(numbers==null||numbers.length==0){ return arr; } int start=0; int end=numbers.length-1.

class Solution { public int findMin(int[] nums) { if(nums==null||nums.length==0){ return -1; } int start=0; int end=nums.length-1; while(start<end){ int mid=start+(end-start)/2; .

class Solution { public int findMin(int[] nums) { if(nums==null||nums.length==0){ return -1; } int start=0; int end=nums.length-1; while(start<=end){ int mid=start+(end-start)/2; .

class Solution { public boolean search(int[] nums, int target) { if(nums.length==0||nums==null){ return false; } int start=0; int end=nums.length-1; while(start<=end){ int mid=start+(.

class Solution { public int findMin(int[] nums) { if(nums==null||nums.length==0){ return -1; } int start=0; int end=nums.length-1; while(start<=end){ int mid=start+(end-start)/2; .

二分法的思想来做这道题目class Solution { public int search(int[] nums, int target) { if(nums==null||nums.length==0){ return -1; } int start=0; int end=nums.length-1; while(start<=end){ int .

public List<String> getRecommendenResult(List<String> picAndVideo,int maxInterval){ List<String> result=new ArrayList<>(); if(picAndVideo==null || picAndVideo.size()==0){ return result; } Queue<String>.

class MyQueue { Stack<Integer> stack1; Stack<Integer> stack2; /** Initialize your data structure here. */ public MyQueue() { stack1=new Stack<>(); stack2=new Stack<>(); } /** Push elem.

核心的思想是：栈+前缀和public int getMax(int[] numbers){ if(numbers.length==0||numbers==nul){ return 0; } Stack<Integer> stack=new Stack<>(); int max=0; //求前缀和的数组 int[] sum=new int[numbers.length+1]; for(int i=1;i&.

最小栈，一定要读懂题目class MinStack { /** initialize your data structure here. */ Stack<Integer> stack; Stack<Integer> minStack; public MinStack() { stack=new Stack(); minStack=new Stack(); } public void pu.