HDOJ 1020 Encoding

Problem Description
Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.

2. If the length of the sub-string is 1, ‘1’ should be ignored.

Input
The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only ‘A’ - ‘Z’ and the length is less than 10000.

Output
For each test case, output the encoded string in a line.

Sample Input
2
ABC
ABBCCC

Sample Output
ABC
A2B3C

题意：按照字符串的顺序，输出字符的个数和字符。
如果字符出现次数为1次，只要输出原字符。
如果输入为：ABBCCCBBB
输出为：A2B3C3B
而不是：A5B3C

import java.util.Scanner;

public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while (t-- > 0) { String strs = sc.next(); // System.out.println(strs+"=strs"); boolean isSee[] = new boolean[strs.length()]; for (int i = 1; i < isSee.length; i++) { isSee[i] = false; } int sum = 0; boolean isLast=false; for (int i = 0; i < strs.length()-1; i++) { if(strs.charAt(i)==strs.charAt(i+1)) { isSee[i]=true; isSee[i+1]=true; sum=sum+1; }else{ sum=sum+1; isSee[i]=false; } if(!isSee[i]){ if(sum==1){ System.out.print(strs.charAt(i)); }else{ System.out.print(""+sum+strs.charAt(i)); } } if(!isSee[i]){ sum=0; } } if(isSee[strs.length()-1]){ System.out.print(""+(sum+1)+strs.charAt(strs.length()-1)); }else{ System.out.print(strs.charAt(strs.length()-1)); } System.out.println(); } }

}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/50943872