LeetCode刷题（C++）——Letter Combinations of a Phone Number（Medium）

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.

思路：第一眼看到这道题觉得不难，只需要利用for循环暴力寻找所有的组合结果就可以了，但是仔细一想发现，我们事先不知道digits的大小，于是就无法确定使用for循环的层数。怎么办呢？嗯，可以用递归来代替for循环。那么递归我们要递归什么呢？当然是要递归digits中的数字啦，然后对每个数字对应的字符进行遍历，来组合所有的可能。

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> letter;if (digits.empty())return letter;vector<string> num = { "abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };string let;dfs(digits, letter, let, 0, num);return letter;}void dfs(string& digits, vector<string>& letter, string& let, int pos, vector<string> num){if (pos == digits.size()){letter.push_back(let);return;}int x = (digits[pos] - '0') - 2;string s = num[x];for (int i = 0; i < s.size();i++){let.push_back(s[i]);dfs(digits, letter, let, pos + 1, num);let.pop_back();}    }};