LeetCode 394. Decode String

February 25, 2020

leetcode 394 solutions

Given an encoded string, return its decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

First Submission

class Solution {
public:
    string decodeString(string s) {
        vector <vector <int>> encoded = findNum(s);
        if (encoded.size() == 0) {
            return s;
        }
        string ret = "";
        int i = 0;
        int j = 0;
        while(i<s.length()) {
            if (s[i] >= '0' && s[i] <= '9') {
                vector<int> token = encoded[j];
                ret += dupString(decodeString(s.substr(token[1]+1,token[2]-token[1]-1)),token[0]);
                j++;
                i = token[2];
            }
            else {
                ret += s[i];
            }
            i++;
        }
        return ret;
    }
    string dupString(string s, int n) {
        string str = "";
        for (int i = 0; i<n; i++) {
            str += s;
        }
        return str;
    }
    
    vector<vector <int>> findNum(string s) { //find [] in string s
        vector<vector <int> >encoded = {}; 
        for(int i = 0; i<s.length(); i++) {
            if (s[i] >= '0' && s[i] <= '9') {
                vector<int> token = {0,0,0};
                bool found = false;
                vector<int> stack = {};
                for(int j = i; j<s.length(); j++) {
                    if (!found && s[j] >= '0' && s[j] <= '9') {
                        token[0] = token[0]*10+s[j]-'0';
                    }
                    if (s[j] == '[') {
                        found = true;
                        stack.push_back(j);
                    }
                    if (s[j] == ']') {
                        if (stack.size() == 1) {
                            token[1] = stack[0];
                            token[2] = j;
                            encoded.push_back(token);
                            stack.pop_back();
                            i = j;
                            break; //found
                        }
                        stack.pop_back();
                    }
                    
                }
            }
        }
        return encoded;
    }
};

Simplified

class Solution {
public:
    string decodeString(string s) {
        int i = 0;
        return decodeString(s,i); //start from index 0;
         // a temporary cannot bind to a non-const reference. Have to declare an int first.
    }
    string decodeString(const string &s, int &i) { // i is starting index
        // Forgot to use reference in parameters
        
        string res = "";
        
        while(i<s.length() && s[i] != ']') {
            if (!isdigit(s[i])) {
                res += s[i++];
            }
            else {
                int n = 0;
                while(i<s.length() && isdigit(s[i])) {
                    n = n*10 + s[i++] - '0';
                }
                i++; // '['
                string str = decodeString(s,i);
                i++; // ']'
                while(n-- > 0) {
                    res += str;
                }
            }
        }
        return res;
    }
};