LeetCode | 0017. Letter Combinations of a Phone Number电话号码的字母组合【Python】
LeetCode 0017. Letter Combinations of a Phone Number电话号码的字母组合【Medium】【Python】【回溯】【DFS】【暴力】
Problem
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.
问题
给定一个仅包含数字 2-9
的字符串,返回所有它能表示的字母组合。
给出数字到字母的映射如下(与电话按键相同)。注意 1 不对应任何字母。
示例:
输入:"23"
输出:["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
说明:
尽管上面的答案是按字典序排列的,但是你可以任意选择答案输出的顺序。
思路
解法一
回溯 DFS
注意要判断 path != '',因为当 digits = '',返回的是 [] 不是 ['']。
Python3代码
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
# solution one: backtracking
kbmaps = {'2': 'abc', '3': 'def', '4': 'ghi', '5': 'jkl', '6': 'mno', '7': 'pqrs', '8': 'tuv', '9': 'wxyz'}
ans = []
self.dfs(digits, 0, ans, '', kbmaps)
return ans
def dfs(self, string, index, ans, path, kbmaps):
if index == len(string):
if path != '': # while digits = '', return [] not ['']
ans.append(path)
return
for i in kbmaps[string[index]]:
self.dfs(string, index + 1, ans, path + i, kbmaps)
解法二
暴力
Python3代码
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
# solution two: brute force search
if digits == "":
return []
d = {'2' : "abc", '3' : "def", '4' : "ghi", '5' : "jkl", '6' : "mno", '7' : "pqrs", '8' : "tuv", '9' : "wxyz"}
ans = ['']
for x in digits:
ans = [y + c for c in d[x] for y in ans]
return ans
代码地址
最怕一生碌碌无为,还说平凡难能可贵。