Leetcode Regular Expression Matching

2018-09-04 11:37:17 浏览数 (1)

题目：Regular Expression Matching

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

代码语言：javascript复制
'.' 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:

代码语言：javascript复制
Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

代码语言：javascript复制
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:

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

Example 4:

代码语言：javascript复制
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:

代码语言：javascript复制
Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

解答：

刷题碰到了正则表达式匹配，之前牛客网做题时候做到过，本来以为同样的思路，结果并没有通过。

以下参考自Regular Expression Matching-sample 48 ms submission

代码语言：javascript复制
class Solution:
    def isMatch(self, s, p):
        m, n = len(s), len(p)
        if n > 0 and p[0] == '*': return False
        
        s, p = s   '#' , p   '#'    # the hashtags mark the end of string/pattern

        prev = [False] * (m 1)
        current = [False] * (m 1)
        prev[-1] = True            # end of pattern matches end of string

        for i in range(n-1,-1,-1):
            if True not in prev: return False     # terminate early
            if p[i] == '*': continue         # ignore * and move to next char in pattern

            elif p[i] != '.' and p[i 1] == '*':
                current[-1] = prev[-1]
                for j in range(m-1,-1,-1):
                    current[j] =  prev[j] or (p[i] == s[j] and current[j 1]) 

            elif p[i] == '.' and p[i 1] == '*':
                current[-1] = prev[-1]
                for j in range(m-1,-1,-1):
                    current[j] = current[j 1] or prev[j]
            elif p[i] == '.':
                for j in range(m-1,-1,-1):
                    current[j] = prev[j 1]
            else:
                for j in range(m-1,-1,-1):
                    current[j] = prev[j 1] and p[i] == s[j]

            prev = current
            current = [False] * (m 1)

        return prev[0]
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