题目表述

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

 示例1

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

示例2

输入：digits = ""
输出：[]

 

解题思路

利用 dfs思想

 

解题过程

 

class Solution {
public://将电话按键 对应的字符串一一映射出来string strA[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};void Combination(string digits,int level,string Combinstr, vector<string>& v){ //递归返回的条件，层数完成后 返回if(level == digits.size()){v.push_back(Combinstr);return ;}int num = digits[level] - '0';string str = strA[num];//去映射关系中取出 对应字符串for(size_t i = 0;i<str.size();i++){Combination(digits,level+1,Combinstr + str[i],v);}}vector<string> letterCombinations(string digits) {//定义一个 接收字符串的 vector       vector<string> v;if(digits.empty())  return v;Combination(digits,0,"",v);return v;}
};

 递归演示