请实现一个函数用来匹配包括'.'和'*'的正则表达式。模式中的字符'.'表示任意一个字符，而'*'表示它前面的字符可以出现任意次（包含0次）。 在本题中，匹配是指字符串的所有字符匹配整个模式。例如，字符串"aaa"与模式"a.a"和"ab*ac*a"匹配，但是与"aa.a"和"ab*a"均不匹配
1 # -*- coding:utf-8 -*- 2 class Solution: 3 # s, pattern都是字符串 4 def match(self, s, pattern): 5 m,n = len(s),len(pattern) 6 dp = [[False for _ in range(n+1)]for _ in range(m+1)] 7 dp[0][0] = True 8 for i in range(1,n+1): 9 if (pattern[i-1] == '*'): 10 dp[0][i] = dp[0][i-2] 11 for i in range(1,m+1): 12 for j in range(1,n+1): 13 if s[i-1] == pattern[j-1] or pattern[j-1] == '.': 14 dp[i][j] = dp[i-1][j-1] 15 elif pattern[j-1] == '*': 16 if pattern[j-2] == s[i-1] or pattern[j-2] == '.': 17 dp[i][j] |= dp[i][j-1] 18 dp[i][j] |= dp[i-1][j] 19 dp[i][j] |= dp[i][j-2] 20 else: 21 dp[i][j] = dp[i][j-2] 22 return dp[m][n] 23 # write code here