[LeetCode] Letter Combinations of a Phone Number 解题报告

最新推荐文章于 2025-07-20 11:42:00 发布

原创最新推荐文章于 2025-07-20 11:42:00 发布 · 154 阅读

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本文介绍了一种使用递归算法来解决电话号码对应的字母组合问题的方法。通过将数字映射为字母，实现所有可能的组合输出。

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.

» Solve this problem

[解题思路]
经典递归题。

[Code]

1:    vector<string> letterCombinations(string digits) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      string trans[] = {"", " ", "abc", "def", "ghi", "jkl",  
5:        "mno", "pqrs", "tuv", "wxyz"};  
6:      vector<string> set;      
7:      string seq;  
8:      Generater(trans, digits, 0, seq, set);  
9:      return set;  
10:    }  
11:    void Generater(string trans[], string& digits,   
12:      int deep, string& result, vector<string>& set)  
13:    {  
14:      if(deep == digits.size())  
15:      {  
16:        set.push_back(result);  
17:        return;  
18:      }  
19:      int curDig = digits[deep] - 48;  
20:      for(int i =0; i < trans[curDig].size(); i++)  
21:      {  
22:        result.push_back(trans[curDig][i]);  
23:        Generater(trans, digits, deep+1, result, set);  
24:        result.resize(result.size() -1);  
25:      }  
26:    }

[注意]
1. Line 19。ACSII码转int要减去48.