Leetcode--1. Two Sum(easy)
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 sameelement twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer,Integer> mp=new HashMap<Integer,Integer>();
int len=nums.length;
for(int i=;i<len;i++){
if(mp.containsKey(nums[i])){
int value=mp.get(nums[i]);
return new int[]{value,i};
}
else{
mp.put(target-nums[i],i);
}
}
return new int[];
}
}
Leetcode--1. Two Sum(easy)的更多相关文章
- 【leetcode】Two Sum (easy)
Given an array of integers, find two numbers such that they add up to a specific target number. The ...
- [leetcode] #112 Path Sum (easy)
原题链接 题意: 给定一个值,求出从树顶到某个叶(没有子节点)有没有一条路径等于该值. 思路: DFS Runtime: 4 ms, faster than 100.00% of C++ class ...
- [LeetCode] 437. Path Sum III_ Easy tag: DFS
You are given a binary tree in which each node contains an integer value. Find the number of paths t ...
- [LeetCode] #1# Two Sum : 数组/哈希表/二分查找/双指针
一. 题目 1. Two SumTotal Accepted: 241484 Total Submissions: 1005339 Difficulty: Easy Given an array of ...
- [array] leetcode - 53. Maximum Subarray - Easy
leetcode - 53. Maximum Subarray - Easy descrition Find the contiguous subarray within an array (cont ...
- LeetCode算法题-Sum of Left Leaves(Java实现)
这是悦乐书的第217次更新,第230篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第85题(顺位题号是404).找到给定二叉树中所有左叶的总和.例如: 二叉树中有两个左叶 ...
- LeetCode算法题-Sum of Two Integers(Java实现)
这是悦乐书的第210次更新,第222篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第78题(顺位题号是371).计算两个整数a和b的总和,但不允许使用运算符+和 - .例 ...
- LeetCode--Array--Two sum (Easy)
1.Two sum (Easy)# Given an array of integers, return indices of the two numbers such that they add u ...
- Java for LeetCode 216 Combination Sum III
Find all possible combinations of k numbers that add up to a number n, given that only numbers from ...
- LeetCode 1 Two Sum 解题报告
LeetCode 1 Two Sum 解题报告 偶然间听见leetcode这个平台,这里面题量也不是很多200多题,打算平时有空在研究生期间就刷完,跟跟多的练习算法的人进行交流思想,一定的ACM算法积 ...
随机推荐
- sql:inner join,left join,right join,full join用法及区别
join的语法: select [字段] from [表名1] inner/left/right/full join [表名2] on [表名1.字段1] <关系运算符> [表名2.字段2 ...
- python string tuple list dict 相互转换的方法
dict = {'name': 'Zara', 'age': 7, 'class': 'First'}# 字典转为字符串,返回:<type 'str'> {'age': 7, 'name' ...
- Mac Terminal
一.简介 二.实用 1)update-apps-using-terminal-mac https://www.maketecheasier.com/update-apps-using-termin ...
- MSDos
一.简介 二.源码 http://www.computerhistory.org/atchm/microsoft-research-license-agreement-msdos-v1-1-v2- ...
- gerrit管理下的git代码提交小技巧
1.提交代码git checkout targetbranch 切换至目标分支git pull origin targetbranch 拉取目标分支最新内容git add 修改文件git commit ...
- 去掉tableView空白区域的分割线
//把多余的分割线去掉 UIView * footerView = [[UIView alloc] initWithFrame:CGRectZero]; self.tableView.tableFoo ...
- hook api实现
https://www.cnblogs.com/findumars/p/8734116.html https://blog.csdn.net/lonelyrains/article/details/2 ...
- Java界面编程—事件监听机制
组件首先要先注册事件处理器,当用户单击组件.移动鼠标或者敲击键盘时都会产生事件(Event),一旦有时间发生,应用程序就会做出对该事件的响应,这些组件就是事件源(Event source). 接受.解 ...
- [Hbase]Hbase技术方案
HBase架构简介 HBase在完全分布式环境下,由Master进程负责管理RegionServers集群的负载均衡以及资源分配,ZooKeeper负责集群元数据的维护并且监控集群的状态以防止单点故障 ...
- where
(二)WHERE //where不单独使用,与match,optional match,start,with搭配 where 与match,optional match 一起用,表示约束 where ...