[Leetcode]10. Regular Expression Matching @python

题目

Implement regular expression matching with support for ‘.’ and ‘*’.

‘.’ Matches any single character.
‘*’ Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:

isMatch(“aa”,”a”) → false
isMatch(“aa”,”aa”) → true
isMatch(“aaa”,”aa”) → false
isMatch(“aa”, “a*”) → true
isMatch(“aa”, “.*”) → true
isMatch(“ab”, “.*”) → true
isMatch(“aab”, “c*a*b”) → true

题目要求

实现正则表达式的匹配。函数isMatch接收字符串s和正则表达式p，正则表达式要支持’.’和’*’操作，其中’.’可以表示任意一个字符，’*’表示0个或多个前面的字符。可以通过例子理解正则表达式。

解题思路

本题参考了南郭子綦的代码。使用动态规划进行求解。dp[i][j] 表示s中前i个字符与p的前j个字符组成的表示式是否匹配。dp[0][0]恒为True，对于i为0的情况，空串可以匹配到p前面的表达式为.* 或c*这种情况（c为任意字符）.

dp[i][j]=⎧⎩⎨⎪⎪⎪⎪⎪⎪dp[i−1][j−1],dp[i][j−1],dp[i][j−2],dp[i−1][j],p[j - 1] == '.' or (s[i - 1][j - 1] == p[j - 1])p[j-1] == '*' and p[j -2] show oncep[j-1] == '*' and p[j -2] doesn't showp[j -1] == '*' and p[j -2] show more than once

代码

class Solution(object):    def isMatch(self, s, p):        """        :type s: str        :type p: str        :rtype: bool        """        dp = [[False for _ in range(len(p) + 1)] for _ in range(len(s) + 1)]        dp[0][0] = True        for j in range(2,len(p) + 1):            if p[j - 1] == '*':                dp[0][j] = dp[0][j - 2]          for i in range(1,len(s) + 1):            for j in range(1,len(p) + 1):                if p[j - 1] == '*':                    dp[i][j] = dp[i][j-1] or dp[i][j-2] or (dp[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))                elif p[j-1] == '.' or s[i-1] == p[j - 1]:                    dp[i][j] = dp[i-1][j-1]        return dp[len(s)][len(p)]