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