Leetcode 394. string decoding

class Solution {
    
public:
    string decodeString(string s) {
    
        string res = "";
        stack <int> nums;
        stack <string> strs;
        int num = 0;
        int len = s.size();
        for(int i = 0; i < len; ++ i)
        {
    
            if(s[i] >= '0' && s[i] <= '9')
            {
    
                num = num * 10 + s[i] - '0';
            }
            else if((s[i] >= 'a' && s[i] <= 'z') ||(s[i] >= 'A' && s[i] <= 'Z'))
            {
    
                res = res + s[i];
            }
            else if(s[i] == '[') // take ‘[’ Press the number before into nums Inside the stack ,  The letter string is pressed into strs Inside the stack 
            {
    
                nums.push(num);
                num = 0;
                strs.push(res); 
                res = "";
            }
            else // encounter ‘]’ when , Operate the matching ‘[’ Characters between , Use the law of distribution 
            {
    
                int times = nums.top();
                nums.pop();
                for(int j = 0; j < times; ++ j)
                    strs.top() += res;
                res = strs.top(); // Then if it's still a letter , Will be added directly to res after , Because they are the same level of operations 
                                  // If it's left bracket ,res Will be pressed into strs Stack , As the operation of the upper layer 
                strs.pop();
            }
        }
        return res;
    }
};

 author ：youlookdeliciousc
 link ：https://leetcode-cn.com/problems/decode-string/solution/ti-jie-czhan-by-youlookdeliciousc/
 source ： Power button （LeetCode）
 The copyright belongs to the author . Commercial reprint please contact the author for authorization , Non-commercial reprint please indicate the source .