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58 lines
2.2 KiB
Python
58 lines
2.2 KiB
Python
#!/usr/bin/env python
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# coding=utf-8
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########################################################################################################
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# Leetcode 17 电话号码的字母组合
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# https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/
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#
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# 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。
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# 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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#
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# 示例:
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# 输入："23"
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# 输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
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#
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# 说明:
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# 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。
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########################################################################################################
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class Solution:
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numMap = {'2': "abc", '3': "def", '4': "ghi", '5': "jkl", '6': "mno", '7': "pqrs", '8': "tuv", '9': "wxyz"}
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def letterCombinations(self, digits):
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"""
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:type digits: str
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:rtype: List[str]
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(Knowledge)
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思路：
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1. 这是一个回溯问题，且没有最优子结构，不能用动态规划，所以可以使用回溯框架用递归方式解决；
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回溯框架套路看这里 => https://labuladong.gitbook.io/algo/di-ling-zhang-bi-du-xi-lie/hui-su-suan-fa-xiang-jie-xiu-ding-ban
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"""
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def _letterCombinations(path, digits, index, result):
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"""
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:type path: str => 当前路径
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:type digits: str => 原数字字符串
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:type index: int => 当前遍历位置
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:type result: List[str] => 存储结果
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"""
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if index == len(digits) - 1:
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for ch in self.numMap[digits[index]]:
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result.append(path + ch)
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return
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for ch in self.numMap[digits[index]]:
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_letterCombinations(path + ch, digits, index + 1, result)
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# 对空字符串进行特判
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if not digits:
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return []
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result = []
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_letterCombinations("", digits, 0, result)
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return result
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if __name__ == '__main__':
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solution = Solution()
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print(solution.letterCombinations("23"), '= ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]')
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