LeetCode_OJ【17】Letter Combinations of a Phone Number

最新推荐文章于 2017-09-11 21:05:04 发布

原创最新推荐文章于 2017-09-11 21:05:04 发布 · 310 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Array #string

LeetCode 专栏收录该内容

41 篇文章

订阅专栏

本文详细阐述了如何通过循环实现数字到字母的组合转换，包括数字对应的字母映射、循环逻辑以及输出结果的整理。

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.

这题没什么特别的地方，乖乖地写循环吧

public class Solution {
    public List<String> letterCombinations(String digits) {
		char[][] letters={{},{},{'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'}};
		List<String> list = new ArrayList<String>();
		for(int i = 0; i < digits.length(); i++){
			if(list.size() == 0){
				for(int p = 0; p < letters[digits.charAt(i) - '0'].length ; p++){
					list.add(String.valueOf(letters[digits.charAt(i) - '0'][p]));
				}
				continue;
			}
			List<String> tmp = new ArrayList<String>();
			for(int j = 0 ; j < letters[digits.charAt(i) - '0'].length ; j++){
				for(String s: list){
					tmp.add(s+letters[digits.charAt(i) - '0'][j]);
				}
			}
			list = tmp;
		}
        return list;
    }
}