leetCode题之正则表达匹配

最新推荐文章于 2024-04-12 10:06:29 发布

jamsan_n

最新推荐文章于 2024-04-12 10:06:29 发布

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分类专栏：软件文章标签： leetcode

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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算法，通过详细的代码示例和测试案例，展示了如何实现正则表达式的完全匹配，特别关注了‘.’和‘*’字符的处理。

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原题

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

即给定字符串s(全部小写), 一个正则表达式p(仅包含’.‘和’*'符号, 字母小写). 输出字符串是否完全匹配.

先写测试代码

import unittest
from main import Solution

class TestMain(unittest.TestCase):
    def setUp(self):
        self.solver = Solution()
        self.cases = [
            {
                "feed": 'abcdefggghi',
                "regexp": 'a..defg*hi',
                "want": True
            },
            {
                "feed": 'some text',
                "regexp": '.*text',
                "want": True
            },
            {
                "feed": 'abcddddffg',
                "regexp": 'n*ab.*g',
                "want": True
            },
            {
                "feed": 'abcddddffg',
                "regexp": 'n.ab.*g',
                "want": False
            },
            {
                "feed": '',
                "regexp": 'n.ab.*g',
                "want": False
            },
            {
                "feed": 'abcdef',
                "regexp": 'abcdefgh',
                "want": False
            },
            {
                "feed": 'aab',
                "regexp": 'c*a*b',
                "want": True
            },
        ]

    def testCase(self):
        for case in self.cases:
            res = self.solver.isMatch(case["feed"], case["regexp"])
            self.assertEqual(res, case["want"], "feed:{}, exp:{}".format(case["feed"], case["regexp"]))

if __name__ == '__main__':
    suite = unittest.TestLoader().loadTestsFromTestCase(TestMain)
    unittest.TextTestRunner(verbosity=2).run(suite)

解法

一般是采用动态规划法, 假设s长度为m, p长度为n, 设置一个dp[m+1][n+1]数组保存判断各种情况的结果。

如果 p[j] == ‘*’ 则 dp[i][j]为True的情况为:

dp[i][j-1] == True; 如 s=‘abcd’, p=‘a.*cd’, i=1, j=2;
dp[i-1][j] 且( s[i-1] == p[j-1] or p[j-1] ==’.’)

如果p[j] != ‘*’ 则dp[i][j]为True的情况为:
dp[i-1][j] and (s[i-1] == p[j] or p[j] == ‘.’)

实际时，将dp[0][0]表示为s=’’, p=’'的情况; 完整代码如下:

class Solution(object):
    def isMatch(self, s, p):
        m = len(s)
        n = len(p)
        dp = [[False] * (n+1) for _ 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] = dp[i][j-2] or ( i>0 and (s[i-1] == p[j-2] or p[j-2] == '.') and dp[i-1][j])
                else:
                    dp[i][j] = i>0 and dp[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.')
        return dp[m][n]

运行

> python3 test.py
testCase (__main__.TestMain) ... ok
----------------------------------------------------------------------
Ran 1 test in 0.001s
OK