LeetCode—— *的字母组合

题目地址：https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/

解题思路：暴力求解，很明显循环的个数根据digits的个数有关，所以我们可以采用递归来进行这种动态的多重循环。

class Solution {
private:
    map<int, char> mp = { {2,'a'}, {3,'d'}, {4,'g'}, {5,'j'}, {6,'m'}, {7,'p'}, {8,'t'}, {9,'w'} };
    vector<string> ans;
    void getAns(string result,int loc,char c,string digits) {
        if (loc == digits.size()) {
            ans.push_back(result + c);
            return;
        }
        int num = digits[loc] - '0' == 7 || digits[loc] - '0' == 9 ? 4 : 3;
        for (int i = 0; i < num; i++)
            getAns(result + c, loc+1, mp[digits[loc] - '0']+i, digits);
    }
public:
    vector<string> letterCombinations(string digits) {
        int i;
        if (digits == "")
            return ans;
        else {
            int num = digits[0] - '0' == 7 || digits[0] - '0' == 9 ? 4 : 3;
            for (i = 0; i < num; i++)
                getAns("", 1, mp[digits[0] - '0'] + i, digits);
        } 
        return ans;
    }
};