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();
        }
    }

}