17. Letter Combinations of a Phone Number

本文介绍了一个算法问题: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Python类解决方案,通过深度优先搜索实现所有可能的组合。

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.




do a dfs search to see all the combinations.

class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        NUMBER_HASH={'2':['a','b','c'],
              '3':['d','e','f'],
              '4':['g','h','i'],
              '5':['j','k','l'],
              '6':['m','n','o'],
              '7':['p','q','r','s'],
              '8':['t','u','v'],
              '9':['w','x','y','z']
             }
        digit_length = len(digits)
        def dfs(number_length, string, result):
            if number_length == digit_length:
                result.append(string)
                return
            for letter in NUMBER_HASH[digits[number_length]]:
                dfs(number_length + 1, string + letter, result)
        result = []
        if len(digits) == 0:
            return result
        dfs(0, '', result)
        return result
        
                


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