LeetCode | #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.

思路：

• 其实就是要遍历所有可能，从每个数字对应的字母中选一个，搭配在一起，如果用暴力枚举的话，还真不好搞而且会超时；
• 递归，关键的是要找规则，规则如下：

//digits-号码串    i-号码串的下标    temp-临时字符串    res-返回值public void dfs(String digits, int i, String temp, List<String> res)

1. 初始条件：从号码串的第一个数字开始，定义临时字符串“”，定义要返回的List，开始递归，dfs(digits, 0, "", res)；
2. 如果号码串遍历完了，就返回List，否则获取该数字对应的字符串，遍历每个字母，添加到临时字符串中，并从号码串的下一个数字开始，递归调用，dfs(digits, i+1, temp2, res)；

import java.util.ArrayList;import java.util.HashMap;import java.util.List;public class LetterCombinations {HashMap<Integer, String> map = new HashMap<Integer, String>(){{put(0, " ");put(1, "");put(2, "abc");put(3, "def");put(4, "ghi");put(5, "jkl");put(6, "mno");put(7, "pqrs");put(8, "tuv");put(9, "wxyz");}};public List<String> letterCombinations(String digits) {List<String> res = new ArrayList<>();dfs(digits, 0, "", res);return res;    }//digits-号码串    i-号码串的下标    temp-临时字符串    res-返回值public void dfs(String digits, int i, String temp, List<String> res){if(i == digits.length()){res.add(temp);return;}String s = map.get(digits.charAt(i)-'0');for(int j=0; j< s.length(); j++){String temp2 = temp +s.charAt(j);dfs(digits, i+1, temp2, res);}}}