[DFS]LeetCode394. Decode String

DFS思想解决，利用]分割，传递i值比较巧妙得判断了每个重复字符串的结束位置。

class Solution {
public:
    
    string decodeString(string s, int &i) {
        string result = "";
        string strone;
        
        while(i < s.length() && s[i] != ']') {
            if(s[i] < '0' || s[i] > '9')
                result += s[i++];
            else{
                int num = 0;
                while(s[i] != '[')
                    num = num * 10 + (s[i++] - '0');
                strone = decodeString(s, ++i);
                i++;

                while(num--)
                    result += strone;
            }
        }
        
        return result;
    }
    
    string decodeString(string s) {
        int i = 0;
        return decodeString(s, i);
    }
};