【leetcode刷题笔记】Letter Combinations of a Phone Number
2014-07-10 19:39
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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.
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
题解:用一个二维数组map保存数字到字母的映射,然后深度优先递归搜索如下的一棵树(以"23"为例,树不完整)
代码如下:
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.
题解:用一个二维数组map保存数字到字母的映射,然后深度优先递归搜索如下的一棵树(以"23"为例,树不完整)
代码如下:
public class Solution { public List<String> letterCombinations(String digits) { ArrayList<String> answer = new ArrayList<String>(); char[][] map = {{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},{'m','n','o'},{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}}; String currentPath = new String(); DeepSearch(map, answer, digits, currentPath); return answer; } public void DeepSearch(char[][] map,List<String> answer,String digits,String currentPath){ if(digits.length() == 0) { String temp = new String(currentPath); answer.add(temp); return; } int now = digits.charAt(0) - '0'; for(int j = 0;j < map[now-2].length;j++){ currentPath += map[now-2][j]; DeepSearch(map, answer, digits.substring(1), currentPath); currentPath = currentPath.substring(0, currentPath.length()-1); } } }
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