17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

代码及思路：

方法1，循环

方法2，Backtrack

class Solution {
public:
    /*vector<string> letterCombinations(string digits) {
       
        vector<string> res(1,"");
        string charmap[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        for(int i = 0; i < digits.size(); i++){
            vector<string> tempres;
            for(int c = 0; c < charmap[digits[i] - '0'].size(); c++){
                for(int j = 0; j < res.size(); j++){
                    tempres.push_back(res[j] + charmap[digits[i] - '0'][c]);
                }
            }
            res = tempres;
        }
        if ((res.size()==1)&&(res[0] == "")) res.clear();// clear()
        return res;
    }*/
    vector<string> letterCombinations(string digits){
        string charmap[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        vector<string> res;
        string local;
        backtracking(charmap, digits, res, 0, local);
        if ((res.size()==1)&&(res[0] == "")) res.clear();// clear()
        //atention: string s, s default ""
        return res;
    }
    void backtracking(const string m[], const string& digits, vector<string>& res, int index, string& local){
        // const string& m[] ,error:declaration of 'm' as array of references 
        if(index == digits.size()){
            res.push_back(local);
        }else{
            for(int c = 0; c < m[digits[index] - '0'].size(); c++){
                local.push_back(m[digits[index] - '0'][c]);
                backtracking(m, digits, res, index + 1, local);
                local.pop_back();
            }
        }
        
    }
};