LeetCode10：Regular Expression Matching

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

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LeetCode：链接

剑指offer同题：剑指Offer_编程题52：正则表达式匹配

1）递归方法

如果P[j+1]!='*'，S[i] == P[j]=>匹配下一位(i+1, j+1)，S[i]!=P[j]=>匹配失败；

如果P[j+1]=='*'，S[i]==P[j]=>匹配下一位(i+1, j)或者(i, j+2)，S[i]!=P[j]=>匹配下一位(i,j+2)。

class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        if s == p:
            return True
        if len(p) > 1 and p[1] == '*':
            if s and (s[0] == p[0] or p[0] == '.'):
                return self.isMatch(s[1:], p) or self.isMatch(s, p[2:])
            else:
                return self.isMatch(s, p[2:])
        elif s and p and (s[0] == p[0] or p[0] == '.'):
            return self.isMatch(s[1:], p[1:])
        return False

2）动态规划

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）。

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)]