2017年8月1日 · You try to print a single character with %s, which is for strings. This leads to undefined behavior -- the correct conversion for a single character is %c. You loop until some "digit character" like '3'. You want to loop until the number 3 instead. Subtract '0' to achieve this.

2014年11月14日 · public class CompressString { String compress (String src) { StringBuilder ret = new StringBuilder (); int p1 = 0;int count = 0;int len = src.length (); while (p1< len && src.char_利用指针完成字符串的压缩例如:aaabbbccc压缩成3a3b3c.

2020年5月9日 · System.out.println(changeStr); } // aabbbcccc -->2a3b4c // 一个函数(方法) 传入一个String类型 并返回一个新的String类型 public static String changeStr(String str) { // 转换---将String类型转换成StringBuffer类型 . StringBuffer stringBuffer = new StringBuffer(str); // 修改---在该字符串后面任意添加一个字符 --用于辅助完成最后一个连续字符的个数及其本身的添加 . stringBuffer.append('_'); // 转换---将该StringBuffer转成一个新的String类型 .

2024年10月12日 · 字符串 "a#2b#3c#4" 应该被解码为 "aabbbcccc"。 请编写一个函数来解码这样的字符串。 这种类型的解码通常涉及到将数字字符替换为其前面的字母按次数重复。

2019年1月23日 · the above question is asked at an interview, the code must accept input like aabbbcccc+++ and should output a@2b@3c@4+@3 based on the number of strings occurrences.

2014年4月6日 · I am trying to compress a String. For example, if the user input is "aaabbcccd" - the output should count the letters and if count is above 1, print the letter, then the number: a3b2c3d. This is what I've come up with but my output in the console is still incorrect. String a = IO.readString(); int aLength = aLength.length(); int repeatedStart = 0;

2014年10月11日 · My task was to implement a method which performs basic string compression by counting sequences of repeating characters. Given "aaabbbccc" it should return "a3b3c3". I have included some sample tests I made up. Please let me know if I have missed any cases. I am looking for the fastest implementation possible with the most concise code.

在采访中会问到上述问题，代码必须像aabbbcccc+++一样接受输入，并且应该根据字符串出现的数量输出a@2b@3c@4+@3。 腾讯云 开发者社区

2011年7月21日 · aabbbcccc这9个字母为一个完整的单位，2003里面有多少个这样的单位，抛出去完整的单位剩下的是5个字母aabbb，最后的是b，看下面的分析数据。 2003÷9=222余5，也就是说有222个AABBBCCCC，剩余5个字母不足AAABBBCCCC，所以从A开始往后数第5个字母，答案 …

2019年8月19日 · I need help on creating a shell script which will take input string and expand as below: Input => output. A2B3C4 => AABBBCCCC. What have you done so far to solve this and what in particular are you stuck at? We're not here to do your job or homework, without sharing what you've tried, this is an off-topic question to me.