package org.lep.leetcode.twosum; import java.util.Arrays; import java.util.HashMap; import java.util.Map; /** * source:https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/#/description * Created by lverpeng on 2017/6/21. * * Given an array…

LeetCode 两数之和 给定一个整数数组,返回两个数字的索引,使它们相加到特定目标. 您可以假设每个输入只有一个解决方案,并且您可能不会两次使用相同的元素. 更多文章查看个人博客 个人博客地址:twoSum 两数之和 [JAVA实现] 方法一 使用双重循环两两相加判断是否等于目标值 public List<String> twoSum2(int[] arr, int sum) { if (arr == null || arr.length == 0) { return new ArrayL…

1.普通数组找两个数,哈希表建立数值和下标的映射,遍历时一边判断一边添加 /* 哇,LeetCode的第一题...啧啧 */ public int [] twoSum(int[] nums, int target) { /* 两个数配合组成target 建立值和下标的映射,遍历数组时一边判断是否有另一半,一边添加新映射到哈希表 */ Map<Integer,Integer> map = new HashMap<>(); for (int i = 0; i < nums.len…

Given an array of integers, 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 must be less than index2. Please note that…

题目: Given an array of integers, 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 must be less than index2. Please note t…

Github 地址:https://github.com/google/sanitizers Wiki 地址:https://github.com/google/sanitizers/wiki/AddressSanitizer 参考: 基本使用:https://blog.csdn.net/c_lazy/article/details/80009627 输出信息的详细解释:https://www.jianshu.com/p/3a2df9b7c353 AddressSanitizer(地址杀菌剂,简…

今天注册了大名鼎鼎的LeetCode,做了一道最简单的算法题目: 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. Example: Given nums = [2, 7, 11, 15], target =…

Description 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. 题意 给定一个数组,和一个定值,找出数组元素的两…

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. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2…

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. Example: Given nums = [2,…

题目: Given an array of integers, 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 must be less than index2. Please note t…

题目: 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. Example: Given nums = [, , , ],…

1). two-sum Given an array of integers, 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 must be less than index2. Pleas…

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. Example: Given nums = [2, 7, 11, 15]…

class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int>result; int i,j,k; map<int,int>h; ;i<nums.size();i++) { if(!h.count(nums[i])) h[nums[i]]=i+; k=h[target-nums[i]]; &&k!=i+) { res…

class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> twoSum1(2); map<int,int> mValueIdex; map<int,int>::iterator it; bool flag=0; for(int i=0;i<nums.size();i++) { //两个一样. if(nums[i]=…

两数之和 (简单) 题目描述 给定一个整数数组和一个目标值,找出数组中和为目标值的两个数: 你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用. 例如: 给定 nums = [2,7,11,15] ,target = 9 因为 nums[0] + nums[1] = 9: 因此返回 [0,1]: v1.0代码如下: 正数.0或重复值通过测试: 异常用例: [-1, -2, -3, -4, -5] -8; 输出 [] /** * @param {number[]} nums * @par…

代码的(判断nums[i]或者是target-nums[i]都可以):…

my answer: ​​​​出现的问题:倒数第二行and i !=s这种情况没有考虑进去,以后要思考全面些…

# -*- coding: utf-8 -*- #!/bin/env python # Python2.7 nums = [2, 4, 7, 0, 12, 6] print sorted(range(len(nums))) # ind是 列表nums 排序后值的索引 ind = sorted(range(len(nums)), key=lambda x: nums[x]) print ind L = [('b',2),('a',1),('e',3),('d',4)] print sorted(L…

Given a non-negative integer c, decide whether there're two integers a and b such that a2 + b2 = c. 只要是two_sum 变形 都可以考虑用hash_set来做. class Solution { public: bool judgeSquareSum(int c) { //这个和两数之和很想 感觉是两数之和的变形 //先求一组两平方和之数 如果这两平方和数都存在的话 就返回true //用has…

最近刷LeetCode比较频繁,就购买了官方的参考电子书 (CleanCodeHandbook),里面有题目的解析和范例源代码,可以省去非常多寻找免费经验分享内容和整理这些资料的时间.惊喜的是,里面的所有源代码都是用java语言写的. 接下来的一段时间里,我会将里面的大部分内容翻译成中文,再加上一些小y自己的解法和扩展内容,以博客的形式发在博客园.我想,这会是一件非常有趣的事情. 以下是翻译的前言部分,第1.4题以及其解析部分. 前言: 嗨,各位刷LeetCode的小伙伴们. 就像你们看到这本书…

一个很常见的问题,找出一个数组中和为给定值的两个数的下标.为了简单一般会注明解只有一个之类的. 最容易想到的方法是循环遍历,这里就不说了. 在JS中比较优雅的方式是利用JS的对象作为hash的方式: var twoSum = function(nums, target) { var hash = {}; var i; for (var i = 0; i < nums.length; i++ ) { if (typeof hash[nums[i]] !== "undefined")…

Design and implement a TwoSum class. It should support the following operations:add and find. add - Add the number to an internal data structure.find - Find if there exists any pair of numbers which sum is equal to the value. For example,add(1); add(…

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 mu…

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. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2…

Array 448.找出数组中所有消失的数 要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间 思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…

利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectangle-in-histo…

开始刷 leetcode, 简单笔记下自己的答案, 目标十一结束之前搞定所有题目. 提高一个要求, 所有的答案执行效率必须要超过 90% 的 python 答题者. 1. Two Sum. class Solution(object): def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ tmp = []…

这几天把之前的设计模式回顾了一遍,整理了一点以前的项目.同学说,打算刷leetcode题目,也勾起了我的兴趣,索性也刷一些题目,再提高一些内功.刚开始进去,leetcode随机分配的题目,直接也就做了,在后来,按顺序做,现在做到了第五题.大概做了如下题目: 1. Two Sum Given an array of integers, find two numbers such that they add up to a specific target number. The function t…