[Algo] 611. Compress String II
Given a string, replace adjacent, repeated characters with the character followed by the number of repeated occurrences.
Assumptions
The string is not null
The characters used in the original string are guaranteed to be ‘a’ - ‘z’
Examples
“abbcccdeee” → “a1b2c3d1e3”
public class Solution {
public String compress(String input) {
// Write your solution here
char[] charArr = input.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < charArr.length; i++) {
char cur = charArr[i];
int count = 1;
while (i + 1 < charArr.length && charArr[i + 1] == charArr[i]) {
count += 1;
i += 1;
}
sb.append(cur).append(count);
}
return sb.toString();
}
}
[Algo] 611. Compress String II的更多相关文章
- [Algo] 175. Decompress String II
Given a string in compressed form, decompress it to the original string. The adjacent repeated chara ...
- [CareerCup] 1.5 Compress String 压缩字符串
1.5 Implement a method to perform basic string compression using the counts of repeated characters. ...
- [LeetCode] 344 Reverse String && 541 Reverse String II
原题地址: 344 Reverse String: https://leetcode.com/problems/reverse-string/description/ 541 Reverse Stri ...
- leetcode 344. Reverse String 、541. Reverse String II 、796. Rotate String
344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) ...
- Reverse Words in a String I & Reverse Words in a String II
Reverse Words in a String I Given an input string, reverse the string word by word. For example,Give ...
- leadcode 541. Reverse String II
package leadcode; /** * 541. Reverse String II * Easy * 199 * 575 * * * Given a string and an intege ...
- NYOJ 1067 Compress String(区间dp)
Compress String 时间限制:2000 ms | 内存限制:65535 KB 难度:3 描写叙述 One day,a beautiful girl ask LYH to help he ...
- LeetCode 541. 反转字符串 II(Reverse String II)
541. 反转字符串 II 541. Reverse String II
- 【leetcode_easy】541. Reverse String II
problem 541. Reverse String II 题意: 给定一个字符串,每隔k个字符翻转这k个字符,剩余的小于k个则全部翻转,否则还是只翻转剩余的前k个字符. solution1: cl ...
随机推荐
- nidlist 问题
错误问题如下: 解决方案: Dao文件 boolean DeleteList(String nidList); 改为: boolean DeleteList(@Param("nidList& ...
- PyCharm下创建并运行我们的第一个Django项目
PyCharm下创建并运行我们的第一个Django项目 准备工作: 假设读者已经安装好python 2x或3x,以及安装好Django,以及Pycharm 1. 创建一个新的工程 第一次运行Pycha ...
- HDU - 6043 KazaQ's Socks(找规律)
题意:有n双袜子,编号1到n,放在衣柜里,每天早晨取衣柜中编号最小的袜子穿,晚上将这双袜子放在篮子里,当篮子里有n-1双袜子时,清洗袜子,直到第二天晚上才洗好,并将洗好的袜子重新放回衣柜. 分析:规律 ...
- 使用Linux系统,是一种什么体验?
导读 同事,从事嵌入式软件开发多年,主要开发环境用的就是linux,最疯狂的一段时间直接把系统装成linux系统,然后在linux下面虚拟一个windows操作系统,主要有些事情必须在windows才 ...
- h5-伸缩布局
1.html代码 <div class="box"> <div class="first">1</div> <div ...
- 虚拟机vmware vmnet8 未识别(转)
原文链接:https://blog.csdn.net/windows_q/article/details/50678646
- 尝试用kotlin做一个app(一)
1.先添加一下anko库 依赖:implementation "org.jetbrains.anko:anko:$anko_version" 版本:ext.anko_version ...
- POJ 2006:Litmus Test 化学公式
Litmus Test Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 1709 Accepted: 897 Descri ...
- kube-proxy详解
KUBE_LOGTOSTDERR="--logtostderr=true"KUBE_LOG_LEVEL="--v=4"NODE_HOSTNAME="- ...
- JS-语句五
for循环的实例 1.九九乘法表: 1*1 1*2 1*3 1*2 2*2 2*3 1*3 2*3 3*3 1*4 2*4 4*3 ...