Leetcode:Letter Combinations of a Phone Number

Description:

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.

分析： 非常有意思的题目，电话按键串可以凑成的所有字符串，因为每个按键对应着3个-4个字符串，故基本也是一个深搜问题，对按键串的每一次按键都

可以对应着某个字符，然后深搜每个按键数字，凑成最后的字符串。

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        map<char,string> phone;
        phone.insert(make_pair('1',""));
        phone.insert(make_pair('2',"abc"));
        phone.insert(make_pair('3',"def"));
        phone.insert(make_pair('4',"ghi"));
        phone.insert(make_pair('5',"jkl"));
        phone.insert(make_pair('6',"mno"));
        phone.insert(make_pair('7',"pqrs"));
        phone.insert(make_pair('8',"tuv"));
        phone.insert(make_pair('9',"wxyz"));
        phone.insert(make_pair('0'," "));
        
        string oneres;
        vector<string> result;
        //if(digits.empty()) return result;
        
        findresult(phone,digits,oneres,result,0);
        return result;
    }
    void findresult(map<char,string> phone,string digits,string& oneres, vector<string>& result, int index)
    {
        if(index==digits.size())
        {
            result.push_back(oneres);
            //oneres.clear();
            return;
        }
        string mapto = (phone.find(digits[index]))->second;
        for(int i=0;i<mapto.size();i++)
        {
            string back = oneres;
            oneres += mapto[i];
            findresult(phone,digits,oneres,result,index+1);
            oneres = back;
        }
    }
};