目录 这是一个对LeetCode题目归类的索引,分类标准参考了July大神的<编程之法>以及LeetCode的tag项.分类可能还不太合理,逐步完善,请见谅~ 题主本人也在一点一点的刷题,这个目录跟着刷的题每天更新~ Hope you enjoy coding! 数组 Array 寻找和为定值的两个数 1.Two Sum 167.Two Sum II 从数组移除元素 26.Remove Duplicates from Sorted Array 27.Remove Element 283.Mov…

目录 Suffix Array Summay 单个字符串问题 两个字符串问题 多个字符串问题 AC-Automaton Summary 求长度为n(2e9)不包含给定字符串的合法串个数 包含至少一个词根长度不超过n(2e9)的字符串个数 Suffix Automaton Summary SAM 的定义 SAM的性质 子串的性质 结束位置 endpos Palindromic Tree(回文自动机) Summary Kmp & ExKmp Summary Manacher Summary Hash…

Codeforces Global Round 17 A. Anti Light's Cell Guessing 坑点:\(n=1,m=1\) 时答案为 \(0\) . 其他情况:当 \(n=1\) 或 \(m=1\) 时,只需要取端点即可.其他情况只需要两个点,也是取两个端点,把离一个点曼哈顿距离为固定值的点连成一条线段,可以发现这两个端点形成的线段只可能有一个交点,即隐藏点. #include <bits/stdc++.h> #define rep(i,a,b) for(int i=a;i…

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 m…

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 m…

You are given an array d1,d2,-,dn consisting of n integer numbers. Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the…

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra mem…

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has enough space (size that is greater or equ…

题意:对于一个有序数组,输出和为target的两个元素的下标.题目保证仅有唯一解. 分析: 法一:二分.枚举第一个元素,二分找另一个元素,时间复杂度O(nlogn),非最优解. class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { int len = numbers.size(); vector<int> ans; for(int i = 0; i <…

414. Third Maximum Number 给一个非空的整数数组,找到这个数组中第三大的值,如果不存在,那么返回最大的值.要求时间复杂度为o(n) 例如: Example 1: Input: [3, 2, 1] Output: 1 Explanation: The third maximum is 1. Example 2: Input: [1, 2] Output: 2 Explanation: The third maximum does not exist, so the maxi…