簡單的回溯法！

class Solution {
public:
    void backTrack(string digits, vector<string> words, string ans, vector<string>& res, int k, int flag[])
    {
        if(k == digits.size())
        {
            res.push_back(ans);
        }
        else
        {
            for(int i=0; i<words[(int
 
)(digits.at(k))-(int)'0'].size(); i++)
            {
                string t = ans;
                ans.push_back(words[(int)(digits.at(k))-(int)'0'].at(i));
                backTrack(digits,words,ans,res,k+1,flag);
                ans = t;//也可以直接ans.pop_back(),而不使用中間變數t
            }
        }
    }
    vector
 
<string> letterCombinations(string digits) {
        string ans;
        int flag[4] = {0,};//0為未用過
        vector<string> words = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        vector<string> res;
        if(digits == "")
            return res;
        backTrack(digits,words,ans,res,
 
0,flag);
        return res;
    }
};