10. Regular Expression Matching

地址：https://leetcode.com/problems/regular-expression-matching/description/

描述：

Given an input string (s) and a pattern (p), 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).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: "." means "zero or more () of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

递归

class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        #递归
        if len(p)==0:
            return len(s)==0
        if (len(p)==1) or (p[1]!='*'):
            if len(s)==0 or (s[0]!=p[0] and p[0]!='.'):
                return False
            return self.isMatch(s[1:], p[1:])
        else:
            i = -1
            length = len(s)
            while i<length and (i<0 or p[0]=='.' or p[0]==s[i]):
                if self.isMatch(s[i+1:], p[2:]):
                    return True
                i += 1
            return False

动态规划

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