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?

class Solution {
public:
    int compress(vector<char>& chars) {
        int tol = 1;    //统计相同字符的数量
        for (int i = 0; i<chars.size(); i++) {
            while (i!=chars.size()-1 && chars[i]==chars[i+1]) {
                chars.erase(chars.begin()+i+1);
                tol++;
            }
            int log = 0; //标记插入数字是1的情况
            int logNum = 0;  //统计插入的数字个数
            while (log==1||tol!=1 && tol>0) {
                chars.insert(chars.begin()+i+1, char(tol%10+'0'));
                tol/=10;
                log = tol==1?tol:0;
                logNum++;
            }
            i+=logNum;
            tol = 1;
        }
        return chars.size();
    }
};