fromtyping import ListclassSolution: def threeSum(self, nums: List[int]) -> List[List[int]]: result=[] n=len(nums) nums.sort()ifnot nums or n<3:returnresult hashset=set()foriinrange(n):forjinrange(i+1,n):forkin
Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. 翻译: 给定一个n个整数的数组S,S中是否存在有a b c三个元素使得a+b+c=0?在S中找到所有的非重复组合,使它们的和为0 注意:答案集不能包含重复的数组。例如:[1,5,-6...
Given an array S of n integers, are there elements a, b, c in S 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. For example, given array S = [-1, 0, 1, 2, -1, -...
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[i]+nums[j]==target:returni,j 分析:代码十分简单...
#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...
public class Solution { public String reverseWords(String s) { // 使用trim()方法去除字符串前后的空格 s = s.trim(); // 使用StringBuilder来存储结果,避免频繁的字符串拼接操作 StringBuilder sb = new StringBuilder(); // 定义一个双指针,分别指向字符串的开头和末尾 int start = 0; int end = s...
Code This branch is9026 commits behinddoocs/leetcode:main. README CC-BY-SA-4.0 license 介绍 本项目包含LeetCode、《剑指 Offer(第 2 版)》、《程序员面试金典(第 6 版)》等题目的相关题解。所有题解均由多种编程语言实现,包括但不限于:Java、Python、C++、JavaScript、C#、Go,日常更新。欢迎Star关注本...
for i in range(n - 1): if (delta := arr[i + 1] - arr[i]) < best: best = delta ans = [[arr[i], arr[i + 1]]] elif delta == best: ans.append([arr[i], arr[i + 1]]) return ans Java class Solution { public List<List<Integer>> minimumAbsDifference(int[] arr) {...
discuss/2921/Share-my-Java-solution-using-dynamic-programming 。 公式还是这个不变 首先定义 P(ij)。 P(i,j)={trues[i,j]是回文串falses[i,j]不是回文串P(i,j)=\begin{cases}true& \text{s[i,j]是回文串}\\false& \text{s[i,j]不是串}\end{cases}(i,j)=⎩⎪⎨⎧truefalse...
Java语言实现代码 class Solution {//这个方法是判断是否为运算符的 private boolean isOperation(String x) {if(x.equals("+") || x.equals("-") || x.equals("*") || x.equals("/")) {return true; } return false; } public int evalRPN(String[] tokens) {Stack stack = new Stack(); ...