17. Letter Combinations of a Phone Number-优快云博客

本文介绍了一种使用回溯算法来生成电话号码所有可能的字母组合的方法。通过将数字映射到对应的字母，针对输入的数字字符串，递归地构建所有可能的组合。

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.

一道回溯题，这里主要是注意vector<string>的初始化方法：

vector<string> maps[10] = {"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};//错误
string maps[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; //正确
vector<string> maps = {"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};//正确

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> result;
        if(digits.empty()) return result;
        vector<string> maps = {"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        result.push_back("");
        for(int i = 0 ; i < digits.size();i++){
            vector<string> temp;
            string str = maps[digits[i] - '0'];
            for(int j = 0 ; j < str.size();j++){
                for(int k = 0; k < result.size();k++){
                    temp.push_back(result[k] + str[j]);
                }
            }
            result = temp;
        }
        return result;
    }
};