LeetCode题解索引

全部最新的题解可以在 我的 github 上找,欢迎 star 和 watch ~ 更新中~~ 说明 这个系列的题解包括用 C++/Java/Python 写的 leetcode 上的算法题目,和 Sql 写的 leetcode 上的数据库题目. 有些题目虽然 AC 了却还没写分析,所以这次就开坑来完成. 链接: 我的 github Leetcode Algorithms Problems Leetcode Database Problems CSDN 题解索引 001.Two_Sum (Med

之前想边做题边写结题报告,发现有点力不从心,而且很多地方也是一知半解,现在重新做题,重新理解.这篇文章主要起一个目录的作用. 128 Longest Consecutive Sequence (Java) [Hard] [Array]

Pascal's Triangle  Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 题意实现一个杨辉三角. 这道题只要注意了边界条件应该很好实现出来,C++实现如下 vector<vector<int>> generate(int

题目来源 https://leetcode.com/problems/next-permutation/ Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible

题目: Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 说明: 1)和1比只是有重复的数字,整体仍采用二分查找 2)方法二 : 实现:  

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 题意 先序遍历二叉树,递归的思路是普通的,能否用迭代呢? 非递归思路:<借助stack>

Reverse Words in a String 考虑几个特殊的情况1.若字符窜s="  "2.字符窜s=“a  b  d     e”3.字符窜s=“ a” class Solution { public: void reverseWords(string &s) { int i; int cas=0; string st[100]; s+=' '; for(i=0;i<s.size();i++) { if(i==0 && s[0]==' ') con

题目: Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, otherwise return -1. You may assume no du

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nod