13 - Letter Combinations of a Phone Number-优快云博客

本文链接：https://blog.youkuaiyun.com/superrobin/article/details/11590993

本文介绍了一种通过递归算法来生成电话号码所有可能的字母组合的方法。给定一个数字字符串，返回该数字字符串可能代表的所有字母组合。文章提供了一个C++实现方案，包括递归函数的详细解释。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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: 递归求下一位数字的遍历，当递归到最后一个数字，将结果push_back到目标字符串数组。

class Solution {
public:
    void qemutation(vector<string> num2ch, vector<string>& res, string digits, string& result, int index)
    {
        
        if(index == digits.length())
        {
            res.push_back(result);
            return;
        }
            
        
        int num = digits[index] - '0';
        string str = num2ch[num];
        for( int i = 0; i < str.length(); i++ )
        {
            string temp = result;
            
            result = result + str.at(i);
            qemutation(num2ch, res, digits, result, index+1);
            
            result = temp;
        }
    }
    
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string>res;
        
        //if(digits == "")
        //    return res;
        
        
        string result = "";
        int index = 0;
        vector<string>num2ch(10,"");
        
        num2ch[0] = " ";
        num2ch[1] = "";
        num2ch[2] = "abc";
        num2ch[3] = "def";
        num2ch[4] = "ghi";
        num2ch[5] = "jkl";
        num2ch[6] = "mno";
        num2ch[7] = "pqrs";
        num2ch[8] = "tuv";
        num2ch[9] = "wxyz";
        
        qemutation( num2ch, res, digits, result, index);
        
        return res;
    }
};