杭电ACM 1020 Encoding java解析

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Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34681 Accepted Submission(s): 15377

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

2ABCABBCCC

Sample Output

ABCA2B3C
解题思路整理：对字符串进行遍历，如果当前的字符与前面的字符相同的话那么我加1记录其个数，如果不相同的时候，排除只有1个的情况，只有在字符重复个数大于1
的时候那么认为是有多个在一起，这是将字符串拼接组成新的串继续下一步操作。
import java.util.ArrayList;import java.util.List;import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int count = scanner.nextInt();int i = 0, countNum = 0, j = 0, j2 = 0;String str ;String str2 = "";char sCopy = 0;List<String> list = new ArrayList<String>();while(i < count){str = scanner.next();sCopy = 0;j = 0;str2 = "";countNum = 0;while(j < str.length()){if(sCopy != str.charAt(j)){if(countNum > 1){str2 = str2 + countNum + sCopy;}else if(j != 0)str2 = str2 + sCopy;countNum = 0;sCopy = str.charAt(j);}if(sCopy == str.charAt(j)){countNum++;}j++;if(j == str.length()){if(countNum > 1){str2 = str2 + countNum + sCopy;}else if(j != 0)str2 = str2 + sCopy;list.add(str2);}}i++;}for(i = 0; i < count; i++){System.out.println(list.get(i));}}}

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