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.

思路：双指针的题，j用来扫描，i用来记录result，注意的是：count如果大于1，比如102，那么1，0，2分别占用一个char的位子；

class Solution {
    public int compress(char[] chars) {
        int i = 0;
        int j = 0;
        while(j < chars.length) {
            char curChar = chars[j];
            int count = 0;
            while(j < chars.length && chars[j] == curChar) {
                count++;
                j++;
            }
            chars[i++] = curChar;
            if(count > 1) {
                String countstr = String.valueOf(count);
                for(int k = 0; k < countstr.length(); k++) {
                    chars[i++] = countstr.charAt(k);
                }
            }
        }
        return i;
    }
}