LeetCode重点题系列之【10. Regular Expression Matching】

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本文介绍了一种使用动态规划解决正则表达式匹配问题的方法，重点讲解了如何处理包含 '.' 和 '*' 的模式，通过具体例子说明了算法的工作原理。

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

This a hard question. We can use dp to solve this problem.

Pattern

string

if pattern[i] equals to ".", or if pattern[i] equals string[i], then we remove this character, so mem[i][j]=mem[i-1][j-1], else if pattern[i] equals to "*", if pattern[i-1] equals to string[i], or if pattern[i-1] equals to ".", mem[i][j]=mem[i][j-2] or mem[i-1][j].

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