443. String Compression

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

After you are done modifying the input array in-place, return the new length of the array.

Follow up:
Could you solve it using only O(1) extra space?

Example 1:

Input:
["a","a","b","b","c","c","c"]

Output:
Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]

Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".

Example 2:

Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

Example 3:

Input:
["a","b","b","b","b","b","b","b","b","b","b","b","b"]

Output:
Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].

Explanation:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.

Note:

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

字符串压缩编码，程序如下所示：

class Solution {
    public int compress(char[] chars) {
        int len = chars.length;
        StringBuilder sb = new StringBuilder();
        int cnt = 1;
        for (int i = 0; i < len; ++ i){
            if (i+1 < len && chars[i+1] != chars[i]){
                sb.append(chars[i]);
                if (cnt != 1){
                    sb.append(cnt);
                }
                cnt = 1;
            }
            else if (i+1 < len && chars[i+1] == chars[i]){
                cnt ++;
            }
        }
        sb.append(chars[len-1]);
        if (cnt != 1){
            sb.append(cnt);
        }
        len = sb.length();
        for (int i = 0; i < len; ++ i){
            chars[i] = sb.charAt(i);
        }
        return sb.length();
    }
}