443. String Compression 字符串压缩

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.

给定一组字符，将其压缩到原位。 压缩后的长度必须始终小于或等于原始数组。 数组的每个元素应该是长度为1的字符（不是int）。 

修改输入数组后，返回数组的新长度。

/**
 * @param {character[]} chars
 * @return {number}
 */
var compress = function(chars) {
    if(!chars){
        return 0
    }
    if(chars.length <= 1){
        return chars.length;
    }
    let res = "";
    let count = chars.length - 1;
    let lastChar = chars[0];
    let repeat = 0;
    for(let i = 1; i <= count; i++){
        let curChar = chars[i];
        if(curChar == lastChar){
            repeat++;
        }
        if(curChar != lastChar || i == count){
            repeat <= 0 ? (res += lastChar) : (res += lastChar + String(repeat+1));
            repeat = 0;
            lastChar = curChar;
        }
    }
    if(chars[chars.length-1] != chars[chars.length-2]){
        res+=chars[chars.length-1];
    }
    res = res.split("");
    for(let i in res){
        chars[i] = res[i];
    }
    chars.length = res.length;
    return res.length;
};

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