leetcode------Letter Combinations of a Phone Number

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标题： Letter Combinations of a Phone Number
通过率： 26.6%
难度： 中等

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.

 

本题描述还是比较清楚的，就是给定一个数字的字符串，电话中每个数字都对用几个字母，那么将这个所以的组合都列出来。即有几个数字就有几个字母，

本题刚开始我也搞不明白，因为是个动态循环，也就是说如果是“123”就是三层循环，“1234”就是四层循环，看到这里也可以看出来这种类似“全排列”的题目一定跟树有关，那么本题就是二叉树的深度遍历过程，每一层的元素是由该数字对应的字母数量决定比如说“234”那么树形就是如下图所示：

那么就是一个递归循环，下面看java代码：

 1 public class Solution {
 2     public List<String> letterCombinations(String digits) {
 3         List<String> result=new ArrayList<String>();
 4         String [] map={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
 5         char [] temp=new char[digits.length()];
 6         climbTree(0,digits,temp,result,map);
 7         return result;
 8     }
 9     public void climbTree(int level,String digits,char [] temp,List<String> result,String[] map){
10         if(level==digits.length()){
11             result.add(String.valueOf(temp));
12             return ;
13         }
14         for(int i=0;i<map[digits.charAt(level)-'0'].length();i++){
15             temp[level]=map[digits.charAt(level)-'0'].charAt(i);
16             climbTree(level+1,digits,temp,result,map);
17         }
18     }
19 }

以下是python代码，摘自github：

 

 1 class Solution:
 2     # @return a list of strings, [s1, s2]
 3 
 4     def letterCombinations(self, digits):
 5         d = ['', '', 'abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']
 6 
 7         def _get(step):
 8             if step == N:
 9                 ret.append(''.join(curr))
10                 return
11             ind = int(digits[step])
12             for c in d[ind]:
13                 curr.append(c)
14                 _get(step + 1)
15                 curr.pop()
16 
17         ret = []
18         curr = []
19         N = len(digits)
20         _get(0)
21         return ret

 

posted @ 2015-02-05 11:27  pku_smile  阅读(261)  评论(0)  编辑  收藏  举报