前言

本文主要提供三种不同的解法，分别是利用python的特性、动态规划、递归方法解决这个问题

使用python正则属性

import reclass Solution2:# @return a booleandef isMatch(self, s, p):return re.match('^' + p + '$', s) != None

使用递归方法

class Solution(object):def isMatch(self, text, pattern):if not pattern:return not textfirst_match = bool(text) and pattern[0] in {text[0], '.'}if len(pattern) >= 2 and pattern[1] == '*':return (self.isMatch(text, pattern[2:]) orfirst_match and self.isMatch(text[1:], pattern))else:return first_match and self.isMatch(text[1:], pattern[1:])

使用动态规划

动态规划与上面的递归的区别就在于用空间换取了时间的复杂度

class Solution1(object):def isMatch(self, text, pattern):memo = {}def dp(i, j):if (i, j) not in memo:if j == len(pattern):ans = i == len(text)else:first_match = i < len(text) and pattern[j] in {text[i], '.'}if j+1 < len(pattern) and pattern[j+1] == '*':ans = dp(i, j+2) or first_match and dp(i+1, j)else:ans = first_match and dp(i+1, j+1)memo[i, j] = ansreturn memo[i, j]return dp(0, 0)

后记

这题在leetcode中算是一道难题，有很多值得品味的地方，后面将继续深入研究

参考文档

https://blog.csdn.net/gatieme/article/details/51049244

https://leetcode.com/problems/regular-expression-matching/solution/#

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LeetCode 10. Regular Expression Matching python特性、动态规划、递归

前言

使用python正则属性

使用递归方法

使用动态规划

后记

参考文档

LeetCode 10. Regular Expression Matching python特性、动态规划、递归相关推荐

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