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.

解题要点：

双指针 i 和 j，i 在原点，由 j 来遍历，当 j 检测到重复的数结束后，先把 i 加入数组（从头开始）并把 j 转换为char，也加进数组。

class Solution {
    public int compress(char[] chars) {
        if(chars.length == 1) return 1;
        int curr = 0, i = 0, j = 1;
        int n = chars.length;
        
        while(i < n){
            while(j < n && chars[i] == chars[j]) j++;

            chars[curr++] = chars[i];

            if(j - i == 1){
                i = j;
               continue; 
            } 

            char[] counts = String.valueOf(j-i).toCharArray();
            for(char c : counts) chars[curr++] = c;
                
            i = j;
        }
        return curr;
    }
}