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”].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

这明显的一道回溯法的题。使用深度优先遍历每一个数字可能相应的字符，每遍历完一个字符回溯回下一个字符。

使用一个字符串记录可能的路径，找到递归退出的条件是遍历完字符串。

唯一要注意的一点是数字到字符之间的映射使用map< int ,string >能够实现，可是直接使用string[]也能够实现。

runtime:0ms

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> result;
        if(digits.size()==0)    
            return result;
        string nums[]={" ","00","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        string path;
        helper(digits,0,path,result,nums);
        return result;

    }

    void helper(string digits,int pos,string &path,vector<string> &result,string * nums)
    {
        if(pos==digits.size())
        {
            result.push_back(path);
            return ;
        }
        string tmp=nums[digits[pos]-‘0‘];
        for(int i=0;i<tmp.size();i++)
        {
            path=path+tmp[i];
            helper(digits,pos+1,path,result,nums);
            path=path.substr(0,path.size()-1);
        }
    }
};