Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicate quadruplets. For exampl…

题目: Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: Elements in a quadruplet (a,b,c,d) must be in non-descending or…

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the r…

3Sum Closest Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution. For example, given arr…

问题: Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicate quadruplets. For ex…

2sum 如果数组是无序的,先排序(n*logn),然后用两个指针i,j,各自指向数组的首尾两端,令i=0,j=n-1,然后i++,j--,逐次判断a[i]+a[j]?=sum,如果某一刻a[i]+a[j]>sum,则要想办法让sum 的值减小,所以此刻i 不动,j--,如果某一刻a[i]+a[j]<sum,则要想办法让sum 的值增大,所以此刻i++,j 不动.所以,数组无序的时候,时间复杂度最终为O(n*logn+n)=O(n*logn),若原数组是有序的,则不需要事先的排序,直接O(n)…

1).2sum 1.题意:找出数组中和为target的所有数对 2.思路:排序数组,然后用两个指针i.j,一前一后,计算两个指针所指内容的和与target的关系,如果小于target,i右移,如果大于,j左移,否则为其中一个解 3.时间复杂度:O(nlgn)+O(n) 4.空间:O(1) 5.代码: void twoSum(vector<int>& nums,int numsSize,int target,vector<vector<int>>& two…

首先明确一点,这个方面的问题设计到的知识点是数组的查找的问题.对于类似的这样的查找操作的具体办法就是三种解决方法: 1.暴力算法,多个for循环,很高的时间复杂度 2.先排序,然后左右夹逼,但是这样会破坏原始数组的下表 3.利用Hash表,直接定位元素,很少的时间复杂度 TwoSum 先来看看最简单的,在一个数组中找两个数的和等于某个数. 这个题目最简简单的方法就是暴力法,所需的时间复杂度是O(n2),但是这是不允许的,所以一个O(n)的方法就是利用Hash表存储数据,这样能够把查找的时间降低下…

一.模板以及题目分类 1.头尾指针向中间逼近 ; ; while (pos1<pos2) { //判断条件 //pos更改条件 if (nums[pos1]<nums[pos2]) pos1++; else pos2--; } 经典的找和的问题都可以从这种思路下手,2数之和,3数之和,还注意要区分是寻找值还是索引(寻找索引则不能排序),是否允许有重复,不允许重复时要怎样避开重复值. 避开重复值的方法,当然,在3sum和4sum中的ij要稍微做修改 && nums[i] == n…

题目: 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Given nums = [2, 7, 1…