LeetCode 17: Letter Combinations of a Phone Number

本文介绍了一个经典的编程问题:给定一个数字字符串,返回所有可能的字母组合。文章提供了一个使用深度优先搜索(DFS)的解决方案,并附带了C++代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Difficulty: 3

Frequency 3


Problem:

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:

class Solution {
public:
    string letters[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> answer;
        vector<char> v_letters;
        DFS (digits, 0, v_letters, answer);
        return answer;
    }
    void DFS(string & digits, int i_level, vector<char> & v_letters, vector<string> & answer)
    {
        if (i_level==digits.size())
        {
            string str_letter;
            for (int i = 0; i<v_letters.size(); i++)
            {
                str_letter += v_letters[i];
            }
            answer.push_back(str_letter);
            return;
        }        
        
        for (int i = 0; i<letters[digits[i_level]-'0'].size(); i++)
        {
            v_letters.push_back(letters[digits[i_level]-'0'][i]);
            DFS(digits, i_level+1, v_letters, answer);
            v_letters.pop_back();
        }
    }
};


Notes:

Just DFS, It is boring.

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值