LeetCode之17----Letter Combinations of a Phone Number

题目：

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.

题目大意：

　　输入一个数字字符串，返回数字对应的每个按钮上所有字母的组合

思路：

　　大体思路是以按键上的数字作为下标，数字键上的字母作为vector的值创建一个vector<string>，然后将传进来的数字字符串替换成数字键上的字母串，然后递归每一个数字字符串中的每一个数字替换之后的字母字符串，每递归一层就在临时字符串的相应位置添加上本次扫描的字符，如果递归到最后一层了则将现在的临时结果字符串添加到结果vector中。

代码：

class Solution {public:    std::vector<std::string> letterCombinations(std::string digits) {        std::vector<std::string> result;        //将手机的每个按键映射到以数字键为下标的数组中        std::vector<std::string> phoneMap = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        const int size = digits.size();        std::vector<std::string> tmp(size);   //将数字字符串替换成每个数字键上的字母                if (size == 0) {            return result;        }                for (int i = 0; i < size; ++i) {            if (digits[i] == '0' || digits[i] == '1') {                return result;            }            tmp[i] = phoneMap[digits[i] - '0'];        }                func(result, digits, tmp, size, 0);                return result;    }private:    void func(std::vector<std::string> &result, std::string &tar, std::vector<std::string> &tmp,const int &len, int pos)    {        if (pos == len) {            result.push_back(tar);  //如果到最后一层了则将其压入结果vector            return;        }        for (int i = 0; i < tmp[pos].size(); ++i) {            tar[pos] = tmp[pos][i];             func(result, tar, tmp, len, pos + 1);        }    }};