10. Regular Expression Matching[H]

https://leetcode.com/problems/regular-expression-matching/

Description

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

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'.' 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:

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Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

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Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

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

Example 4:

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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:

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Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Solution

https://leetcode.com/problems/regular-expression-matching/solution/

https://leetcode.com/discuss/93024/easy-dp-java-solution-with-detailed-explanation

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class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        if s == p:
            return True
        m, n = len(s), len(p)
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True
        for j in range(1, n):
            if p[j] == '*' and dp[0][j - 1]:
                dp[0][j + 1] = True
        # print dp
        for i in range(m):
            for j in range(n):
                if p[j] == '.' or p[j] == s[i]:
                    dp[i + 1][j + 1] = dp[i][j]
                elif p[j] == '*':
                    if p[j - 1] != s[i] and p[j - 1] != '.':
                        dp[i + 1][j + 1] = dp[i + 1][j - 1]
                    else:
                        dp[i + 1][j + 1] = dp[i + 1][j] or dp[i][j + 1] or dp[i + 1][j - 1]
        return dp[m][n]

44. Wildcard Matching[H]

https://leetcode.com/problems/wildcard-matching/

Description

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

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'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

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:

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Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

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Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

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Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

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Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

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Input:
s = "acdcb"
p = "a*c?b"
Output: false

Solution

1