python回溯求解电话号码组合

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。答案可以按任意顺序返回。

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。
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输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
示例 2：输入：digits = ""
输出：[]
示例 3：输入：digits = "2"
输出：["a","b","c"]

暴力枚举法

class Solution:def letterCombinations(self, digits: str):ans, dct = [""], {'2': "abc", '3': "def", '4': "ghi", '5': "jkl", '6': "mno", '7': "pqrs", '8': "tuv", '9': "wxyz"}for c in digits:ans = [s1 + s2 for s1 in ans for s2 in dct[c]]return ans if ans[0] else []

回溯法

class Solution:def letterCombinations(self, digits: str):if not digits:return list()phoneMap = {"2": "abc","3": "def","4": "ghi","5": "jkl","6": "mno","7": "pqrs","8": "tuv","9": "wxyz",}def backtrack(index: int):if index == len(digits):combinations.append("".join(combination))else:digit = digits[index]for letter in phoneMap[digit]:combination.append(letter)backtrack(index + 1)combination.pop()combination = list()combinations = list()backtrack(0)return combinations