Problem: 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)…
题目难度:Medium 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. 翻译: 给定一个n个整数的数组S,S…
题目难度:Medium 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. 翻译: 给定一个n个整数的数组S,S…
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 15: 3Sumhttps://oj.leetcode.com/problems/3sum/ 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…
以下是这段时间抽时间刷的前5题,都是自己想的解法,或许不是最优解,只是整理下,方便日后优化提升 1. Two Sum: class Solution: # @return a tuple, (index1, index2) def twoSum(self, num, target): dict = {} for i in xrange(len(num)): if dict.get(target-num[i], None) == None: dict[num[i]] = i else: retur…
Problem: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…
之前写过一篇,这是第二篇.上一篇用了多种编程语言来做,这一次是以学算法为主,所以打算都用python来完成. 4. Median of Two Sorted Arrays There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log…
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 so…
题目如下: Python代码: def threeSum(self, nums): res = [] nums.sort() for i in xrange(len(nums)-2): if i > 0 and nums[i] == nums[i-1]: continue l, r = i+1, len(nums)-1 while l < r: s = nums[i] + nums[l] + nums[r] if s < 0: l +=1 elif s > 0: r -= 1 el…