题目 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 = [-…
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, 三数之和,题解,leetcode, 力扣,Python, C++, Java 题目地址: https://leetcode.com/problems/3sum/description/ 题目描述: Given an array nums of n integers, are there elements a, b, c in nums suc…
一天一道LeetCode系列 (一)题目 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. For example, given array S = {-1 0 1 2 -1 -4}, A solution set is: (-1, 0,…
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/ 015. 3Sum 问题 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…
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…
思考的方向不对,即使用了多于别人几倍的时间,也不一定能够达到终点. 我的错误的想法(可以跳过):在leetcode上面做的第四道题,走路一个很大的弯路,收到之前做过的 Container With Most Water 的思路的影响,自己也想到了可以使用两个指针从左右遍历数组, 然而自己在这里走偏了,上来的第一个想法就是可以将整个解集合分为四种情况:全零,一正一负一零,两正一负,两负一正.这样整个程序就大致分为了四块,出现这种想法是由于我在第一层循环上面使用了两个指针向中间遍历,为内循环里面…
题目: 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: Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c) The s…
题目描述: 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. 解题分析: 这道题注意一下几点即可: 1,先固定前两个数的位置,移动第三个数,这样的查找方式不会有数字组合的遗漏: 2,要考虑到数组元素有重复的情况下的处理. 3,若先为数组排…
第一次解: res = [] nums.sort() if len(nums)<3:return [] for i in range(len(nums)-2): left = i+1 right = len(nums)-1 while left < right: val = nums[i]+nums[left]+nums[right] if val==0 and [nums[i],nums[left],nums[right]] not in res: res.append([nums[i],n…