Daily LeetCode 17. Letter Combinations of a Phone Number

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

Medium

问题描述

Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

Example:

Input: "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.

思路及代码

九宫格键盘每个数字都对应着一些字母，给一个数字串，求出所有数字对应的字母之间的组合。

首先我们建立一个字典保存每个数字对应的字母，初始化res数组为''，遍历字符串digits

在遍历过程中，保存当前数字对应的字母串，初始化数组保存当前的字母组合结果，用一个二重循环来生成当前的字母组合。

代码如下：

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        dict = {
            '2': 'abc',
            '3': 'def',
            '4': 'ghi',
            '5': 'jkl',
            '6': 'mno',
            '7': 'pqrs',
            '8': 'tuv',
            '9': 'wxyz'
        }

        res = [''] if digits else []

        for digit in digits:
            current_string = dict[digit]
            current_res = []
            for comb in res:
                for letter in current_string:
                    current_res.append(comb + letter)
            res = current_res

        return res

这个问题还可以用递归来解

def letterCombinations(self, digits):
    if not digits:
        return []
    dic = {"2":"abc", "3":"def", "4":"ghi", "5":"jkl", "6":"mno", "7":"pqrs", "8":"tuv", "9":"wxyz"}
    res = []
    self.dfs(digits, dic, 0, "", res)
    return res
    
def dfs(self, digits, dic, index, path, res):
    if len(path) == len(digits):
        res.append(path)
        return 
    for i in range(index, len(digits)):
        for j in dic[digits[i]]:
            self.dfs(digits, dic, i+1, path+j, res)