Problem Description
Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:
Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
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();
}
}
}