LeetCode练习10：Regular Expression Matching

题目

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

s	p	Output	Explanation
‘aa’	‘a’	False	“a” does not match the entire string “aa”.
‘aa’	‘a*’	True	‘*’ means zero or more of the precedeng element, ‘a’. Therefore, by repeating ‘a’ once, it becomes “aa”.
“ab”	“.*”	True	“.” means “zero or more () of any character (.)”.
“aab”	“c*a*b”	True	c can be repeated 0 times, a can be repeated 1 time. Therefore it matches “aab”.
“mississippi”	“mis*is*p*.”	False	-

分析

题目的输入为：字符串$s$，匹配模式$p$
输出为：匹配结果，True 或者 False

本题可以使用动态规划的方式进行求解。如果定义$P[i][j]$为True时，表示$s[0,1,\cdots,i)$与$p[0, \cdots, j)$匹配上，则递推公式：
1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
2. P[i][j] = P[i][j - 2], if p[j - 1] == '*' 此时$p[j-2]$重复0次；
3. P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'),如果p[j - 1] == '*'且$p[j-2]$至少重复一次；

解答

class Solution:
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        m, n = len(s), len(p)
        dp = [[False]*(n+1) for i in range(m+1)]
        dp[0][0] = True

        for i in range(0, m+1):
            for j in range(1, n+1):
                if p[j-1] != '*':
                    dp[i][j] = i > 0 and dp[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.')
                else:
                    dp[i][j] = dp[i][j-2] or (i > 0 and dp[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))

        return dp[m][n]

此外本题也有另一种解法，即两个方向的匹配动态规划解法。