[leetcode] 17 letter combinations of a phone number

Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.
Example:
Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

这题就是 简单的组合。 然而排列组合我就只会递归的写法。。。。-___-
在每个位置上， 我们都对每一种可能递归展开。
比如_Combinations("23",0,"") 也就是 "23"[0] 位置上有“abc”三种可能，那我们就递归

1. _Combinations("23",1,"a")
   //"23"[1] 的位置上有“def” 3中可能
   1.1 _Combinations("23",2,"ad")
   1.2 _Combinations("23",2,"ae")
   1.3 _Combinations("23",2,"af")
2. _Combinations("23",1,"b")
   ........
3. _Combinations("23",1,"c")
   ........

递归出口也很简单 .
代码如下。

#include <unordered_map>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
class Solution {
    
public:
    vector<string> letterCombinations(string digits) {
        
        if (digits.empty()){
            return ans;
        }
        _Combinations(digits,0,"");
        return ans;
    }
    
private:
    unordered_map<char,string> _up={
        {'2',"abc"},
        {'3',"def"},
        {'4',"ghi"},
        {'5',"jkl"},
        {'6',"mno"},
        {'7',"pqrs"},
        {'8',"tuv"},
        {'9',"wxyz"}
    };

    vector<string> ans;

    void _Combinations(string& digits, int cur, string comb){
        
        if ( cur >= digits.size()){
            ans.push_back(comb);
            return;
        }
        for(auto le: _up[digits[cur]]){
           _Combinations(digits,cur+1,comb + le);
        }
    }   
};

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