LeetCode – Refresh – Letter Combination of a Phone Number

This is just a combination. Use hashtable to hold the number ==> chars

notes:

1. check corner case : input is empty, do not return vector contains empty string.

2. mapping is char to string. Do not use index to string mapping

 

class Solution {
private:
    unordered_map<char, string> dict;
public:
    void getComb(vector<string> &result, string s, string current, int index) {
        if (current.size() == s.size()) {
            result.push_back(current);
            return;
        }
        for (int i = 0; i < dict[s[index]].size(); i++) {
            getComb(result, s, current + dict[s[index]][i], index+1);
        }
    }
    vector<string> letterCombinations(string digits) {
        vector<string> result;
        if (digits.size() == 0) return result;
        dict['2'] = "abc";
        dict['3'] = "def";
        dict['4'] = "ghi";
        dict['5'] = "jkl";
        dict['6'] = "mno";
        dict['7'] = "pqrs";
        dict['8'] = "tuv";
        dict['9'] = "wxyz";
        getComb(result, digits, "", 0);
        return result;
    }
};