Letter Combinations of a Phone Number 解答

最新推荐文章于 2025-09-10 22:21:01 发布

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最新推荐文章于 2025-09-10 22:21:01 发布

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文章标签： java

原文链接：http://www.cnblogs.com/ireneyanglan/p/4888458.html

本文介绍了一种算法，用于根据电话按键对应的字母映射，找出所有可能的字母组合。输入为一串数字，输出为所有可能的字符串组合。

Question

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.

Solution

We can draw out the solution space tree and then dfs traverse this tree.

For example, input is "23"

　　　　　　　　' '

　　　　　　/　　|　　\

　　　　　'a' 　　'b'　　'c'

　　　/　|　\　/　|　\　/　|　\

　　'd' 'e' 'f''d' 'e' 'f' 'd' 'e' 'f'

Two java tricks:

Convert String to int:　　Integer.parseInt(string);

Convert char to int:　　 Character.getNumericValue(element.charAt(2));

 1 public class Solution {
 2     public static final char[][] telephone = {
 3         {'a','b','c',' '},
 4         {'d','e','f',' '},
 5         {'g','h','i',' '},
 6         {'j','k','l',' '},
 7         {'m','n','o',' '},
 8         {'p','q','r','s'},
 9         {'t','u','v',' '},
10         {'w','x','y','z'}
11     };
12     
13     public List<String> letterCombinations(String digits) {
14         List<String> result = new ArrayList<String>();
15         if (digits == null || digits.length() < 1)
16             return result;
17         dfs(digits, 0, new ArrayList<Character>(), result);
18         return result;
19     }
20     
21     private void dfs(String digits, int start, List<Character> list, List<String> result) {
22         if (start < 0)
23             return;
24         if (start >= digits.length()) {
25             int size = list.size();
26             StringBuilder sb = new StringBuilder(size);
27             for (char tmpChar : list)
28                 sb.append(tmpChar);
29             result.add(sb.toString());
30             return;
31         }
32         // Convert char to int
33         int index = Character.getNumericValue(digits.charAt(start));
34         if (index < 2 || index > 9)
35             return;
36         char[] chars = telephone[index - 2];
37         for (char tmpChar : chars) {
38             if (tmpChar != ' ') {
39                 list.add(tmpChar);
40                 dfs(digits, start + 1, list, result);
41                 list.remove(list.size() - 1);
42             }
43         }
44     }
45 }