原本我的做法是重建一個string，但看到別人是直接用數的，直接甘拜下風~~~

ref:https://leetcode.com/problems/decoded-string-at-index/solutions/4094647/97-47-reverse-traversal/?envType=daily-question&envId=2023-09-27

class Solution {
public:
    string decodeAtIndex(string s, int k) {
        long long length = 0;
        int i = 0;

        while(length < k){
            if(isdigit(s[i])){
                length *= s[i] - '0'; 
                cout << length;
            }else{
                length++;
            }
            i++;
        }
        for(int j = i - 1; j >= 0; j--){
            if(isdigit(s[j])){
                length /= s[j] - '0';
                k %= length;
            }
            else{
                if(k == 0 || k == length){
                    return string(1, s[j]);
                }
                length--;
            }
        }
        return "";
    }

};