HDU 1020 Encoding

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本文深入探讨了一种字符串压缩算法，该算法通过对连续重复字符进行计数并替换为计数和字符的方式实现字符串的有效压缩。文章提供了详细的算法描述，输入输出说明，以及样例输入输出，同时附带了Java实现代码，帮助读者理解并实践该压缩算法。

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57816 Accepted Submission(s): 25811

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

Author

ZHANG Zheng

这道题不是让求排完序后的字符串的连续字符个数，而是未排序的字符串的连续字符个数，排序显得画蛇添足了。

附代码：

import java.util.Scanner;
public class Main{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int c = sc .nextInt();
		while(c-->0) {
			String a = sc.next(),s1="";
			char []s = a.toCharArray();
		    s1+=s[0];
			for(int i=1; i<s.length; ++i) {
				if(s[i]!=s1.charAt(s1.length()-1))
					s1+=s[i];
			}
			int []aa = new int[s1.length()];
			for(int i=0; i<s1.length();++i) {
				aa[i]=0;
			}
			for(int i=0,j=0; i<s.length; ++i) {
				if(s[i]==s1.charAt(j)) {
					aa[j]++;
				}
				else {
					j++; 
					aa[j]++;
				}
			}
			String s3="";
			for(int i=0; i<s1.length(); ++i) {
				if(aa[i]==1)
				s3+=s1.charAt(i);
				else
				s3=s3+aa[i]+s1.charAt(i);
			}
		    System.out.println(s3);
//			System.out.println(s1);
//			for(int i=0; i<s1.length();++i) {
//				System.out.println(aa[i]);
//			}
			
		}
		sc.close();
	}
}