LeetCode String Compression

原题链接在这里：https://leetcode.com/problems/string-compression/description/

题目：

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.

题解：

计数连续相同char的个数加在后面.

Time Complexity: O(chars.length). Space: O(1).

AC Java:

class Solution {
    public int compress(char[] chars) {
        if(chars == null || chars.length == 0){
            return 0;
        }
        
        int mark = 0;
        int write = 0;
        for(int i = 0; i<chars.length; i++){
            if(i+1==chars.length || chars[i]!=chars[i+1]){
                chars[write++] = chars[mark];
                if(mark<i){
                    for(char c : (""+(i-mark+1)).toCharArray()){
                        chars[write++] = c;
                    }
                }
                mark = i+1;
            }
        }
        
        return write;
    }
}