[leetcode] 17. Letter Combinations of a Phone Number

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.

这道题是求出一串数字按键可能代表的字符串，缅怀一下Nokia，题目难度为Medium。

题目本身比较简单，典型的递归回溯题目，具体就不详细说明了，对回溯法还不太熟悉的同学请查看八皇后题目（传送门），具体代码：

class Solution {    void getLetters(vector<string>& rst, string curRst, const string& digits, int pos) {        if(pos == digits.size()) {            rst.push_back(curRst);            return;        }                vector<char> letter(4, 'a');        int cnt = 0;        switch(digits[pos]) {            case '2': {                cnt = 3;                letter[0] = 'a';                letter[1] = 'b';                letter[2] = 'c';                break;            }            case '3': {                cnt = 3;                letter[0] = 'd';                letter[1] = 'e';                letter[2] = 'f';                break;            }            case '4': {                cnt = 3;                letter[0] = 'g';                letter[1] = 'h';                letter[2] = 'i';                break;            }            case '5': {                cnt = 3;                letter[0] = 'j';                letter[1] = 'k';                letter[2] = 'l';                break;            }            case '6': {                cnt = 3;                letter[0] = 'm';                letter[1] = 'n';                letter[2] = 'o';                break;            }            case '7': {                cnt = 4;                letter[0] = 'p';                letter[1] = 'q';                letter[2] = 'r';                letter[3] = 's';                break;            }            case '8': {                cnt = 3;                letter[0] = 't';                letter[1] = 'u';                letter[2] = 'v';                break;            }            case '9': {                cnt = 4;                letter[0] = 'w';                letter[1] = 'x';                letter[2] = 'y';                letter[3] = 'z';                break;            }            default: break;        }        for(int i=0; i<cnt; i++) {            getLetters(rst, curRst+letter[i], digits, pos+1);        }    }public:    vector<string> letterCombinations(string digits) {        vector<string> rst;        if(digits.empty()) return rst;        getLetters(rst, "", digits, 0);        return rst;    }};