class Solution
{
public:
bool backspaceCompare(string S, string T)
{
int szs=S.size();
int szt=T.size();
int ends=,endt=;
for(int i=;i<szs;i++) //get the result string of S
{
if(S[i]=='#')
{
if(ends>)
ends--;
}
else
S[ends++]=S[i];
} for(int j=;j<szt;j++) //get the result string of T
{
if(T[j]=='#')
{
if(endt>)
endt--;
}
else
T[endt++]=T[j];
} if(ends!=endt)
return false;
for(int k=;k<ends;k++) //compare two result
{
if(S[k]!=T[k])
return false;
}
return true;
}
};

844. Backspace String Compare的更多相关文章

  1. 【Leetcode_easy】844. Backspace String Compare

    problem 844. Backspace String Compare solution1: class Solution { public: bool backspaceCompare(stri ...

  2. 844. Backspace String Compare判断删除后的结果是否相等

    [抄题]: Given two strings S and T, return if they are equal when both are typed into empty text editor ...

  3. [LeetCode] 844. Backspace String Compare 退格字符串比较

    Given two strings S and T, return if they are equal when both are typed into empty text editors. # m ...

  4. [LeetCode&Python] Problem 844. Backspace String Compare

    Given two strings S and T, return if they are equal when both are typed into empty text editors. # m ...

  5. 【LeetCode】844. Backspace String Compare 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 字符串切片 栈 日期 题目地址:https://le ...

  6. [LeetCode] Backspace String Compare 退格字符串比较

    Given two strings S and T, return if they are equal when both are typed into empty text editors. # m ...

  7. [Swift]LeetCode844. 比较含退格的字符串 | Backspace String Compare

    Given two strings S and T, return if they are equal when both are typed into empty text editors. # m ...

  8. LeetCode - Backspace String Compare

    Given two strings S and T, return if they are equal when both are typed into empty text editors. # m ...

  9. leetcode-844 Backspace String Compare

    Given two strings S and T, return if they are equal when both are typed into empty text editors. # m ...

随机推荐

  1. LibreOJ 6281 数列分块入门5

    题目链接:https://loj.ac/problem/6281 参考博客:https://blog.csdn.net/qq_36038511/article/details/79725027 我一开 ...

  2. 【转】iOS 自动化性能采集

      前言 对于iOS总体生态是比较封闭的,相比Android没有像adb这种可以查看内存.cpu的命令.在日常做性能测试,需要借助xcode中instruments查看内存.cpu等数据. 但是借助i ...

  3. UVa 548 Tree(二叉树最短路径)

    You are to determine the value of the leaf node in a given binary tree that is the terminal node of ...

  4. python脚本netifaces模块的调用

    # vim get_ip.py # -*- coding: utf- -*- #complete local network card IP #need install netifaces modem ...

  5. JavaScript 中 如何判断一个元素是否在一个数组中

    <script type="text/javascript"> var arrList=['12','qw','q','v','d','t']; console.log ...

  6. Jmeter常用脚本开发之Beanshell Sampler

    Beanshell Sampler Beanshell介绍:是一种完全符合java语法规范的脚本语言,且又拥有自己的一些语法和方法:是一种松散类型的脚本语言:它执行标准java语句和表达式,另外它还包 ...

  7. C# 一个数组未赋值引发的错误

    在电脑前又是一天,后来脑子也糊里糊涂,可能是基础还不牢固,设置断点,找了找问题才发现数组定义出了问题, 我是这样定义数组的,string[] auths ; string auths=new stri ...

  8. Linux pip安装使用详解

    简介 pip是Python有它自己的包管理工具,与yum和apt-get相似. 安装步骤: 1.下载get-pip.py:https://bootstrap.pypa.io/get-pip.py 2. ...

  9. Linux下查看系统启动 、运行以及安装时间

    1.uptime命令  例子显示已启动13天 当前用户一个 [root@localhost-live version-build-related]# uptime :: up days, :, use ...

  10. CH#56C 异象石

    一道LCA 原题链接 先跑一边\(dfs\),求出每个节点的时间戳,如果我们将有异象石的节点按时间戳从小到大的顺序排列,累加相邻两节点之间的距离(首尾相邻),会发现总和就是答案的两倍. 于是我们只需要 ...