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

Solution：Iterate

思路：
Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

// start .. i .. end
class Solution {
    public int compress(char[] chars) {
        int start = 0;
        for (int i = 0; i < chars.length; i++) {
            int end = i;
            while (end < chars.length && chars[end] == chars[i]) {
                end++;
            }
            if (end == i + 1) {
                chars[start++] = chars[i];
            } else {
                chars[start++] = chars[i];
                char[] count = String.valueOf(end - i).toCharArray();
                for (int l = 0; l < count.length; l++) {
                    chars[start++] = count[l];
                }
            }
            i = end - 1;
        }
        return start;
    }
}