classSolution:deftwoSum(self,nums,target):""" :type nums: List[int] :type target: int :rtype: List[int] """#对nums每个元素循环foriinrange(len(nums)):#从nums[i]的下一个开始找forjinrange(i+1,len(nums)):#如果找到了就返回值ifnums[
#TitleSolutionDifficultyAnalysis 306 Additive Number Java Medium 305 Number of Islands II Java Hard 304 Range Sum Query 2D - Immutable Java Medium 303 Range Sum Query - Immutable Java Easy 302 Smallest Rectangle Enclosing Black Pixels Java Medium 300 Longest Increasing Subsequence Java Medium...
Elements in a triplet (a,b,c) must be in non-descending order. (ie,a≤b≤c) The solution set must not contain duplicate triplets. For example, given array S = {-1 0 1 2 -1 -4}, A solution set is: (-1, 0, 1) (-1, -1, 2) 题解: 3 Sum是two Sum的变种,可以利用two ...
的情况时,其实只用到了 4 , 3 长度的情况,而长度为 1 2 的子串情况其实已经不需要了。但是由于我们并不是用 P 数组的下标进行的循环,没有想到优化的方法。 之后看到了另一种动态规划的思路 https://leetcode.com/problems/longest-palindromic-substringdiscuss/2921/Share-my-Java-solution-using-...
Solution3. Java /*// Definition for a Node.class Node {public int val;public List<Node> children;public Node() {}public Node(int _val,List<Node> _children) {val = _val;children = _children;}};*/classSolution{publicList<Integer>postorder(Noderoot){List<Integer>result=newArrayList<>();...
1. Description: Notes: 2. Examples: 3.Solutions: 1/**2* Created by sheepcore on 2018-11-113*/4classSolution {5publicString[] reorderLogFiles(String[] logs) {6Comparator<String> myComp =newComparator<String>() {7@Override8publicintcompare(String s1, String s2) {9ints1si = s1.indexOf...
public class Solution { public String longestCommonPrefix(String[] strs) { if (strs == null || strs.length == 0) { return ""; } // 假设第一个字符串是最长公共前缀的初始值 String prefix = strs[0]; // 遍历字符串数组 for (int i = 1; i < strs.length; i++) { // 如果当...
Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. ...
leetcode刷题(3) Java代码实现 class Solution {public boolean isValid(String s) {Stack stack = new Stack(); for (int i = 0; i < s.length(); i++) {char ch = s.charAt(i); if(ch == '(' || ch == '{' || ch == '[') {stack.push(ch); ...
1. 2. 3. 4. 5. 6. 7. 8. 9. 解题步骤 ①递归树 (绿色箭头上面的是路径,红色框[]则为结果,黄色框为选择列表) 从上图看出,组合问题和子集问题一样,1,2 和 2,1 `是同一个组合,因此 需要引入start参数标识,每个状态中选择列表的起始位置。另外,每个状态还需要一个 sum 变量,来记录当前路径的和,...