我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.gitbooks.io/leetbook/ 16. 3Sum Closest [M] Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target.…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 16: 3Sum Closesthttps://oj.leetcode.com/problems/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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:3sum, three sum, 三数之和,题解,leetcode, 力扣,Python, C++, Java 题目地址: https://leetcode.com/problems/3sum-closest/description/ 题目描述: Given an array nums of n integers and an integer targe…

一天一道LeetCode系列 (一)题目: 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 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 array S = {-1 2…

题目: 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 array S = {-…

1. 原题链接 https://leetcode.com/problems/3sum-closest/description/ 2. 题目要求 数组S = nums[n]包含n个整数,找出S中三个整数a,b,c,使得a+b+c=sum,sum最接近给定的目标整数target,返回sum. 3. 解题思路 采用与第15题相同的思路(第15题链接),不过要引入两个整型变量min和result. min用来保存当前sum和target的最小差值 result用来保存最小差值时的sum,最终的resul…

LeetCode 16. 3Sum Closest(最接近的三数之和)…

16. 3Sum Closest Medium 131696FavoriteShare Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have e…

n数求和,固定n-2个数,最后两个数在连续区间内一左一右根据当前求和与目标值比较移动,如果sum<target,移动较小数,否则,移动较大数 重复数处理: 使i为左至右第一个不重复数:while(i != 0 && i<n-2 && a[i] == a[i-1]) i++; 使r为右至左第一个不重复数(原序最后一个):while(r>i && r+1<n && a[r] == a[r+1]) r--; 15. 3Sum…