regular-expression-matching.py

# -*- coding:utf-8 -*-


# 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.
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# The matching should cover the entire input string (not partial).
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# Note:
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# 	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 *.
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# Example 1:
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# Input:
# s = "aa"
# p = "a"
# Output: false
# Explanation: "a" does not match the entire string "aa".
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# 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".
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# Example 3:
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# Input:
# s = "ab"
# p = ".*"
# Output: true
# Explanation: ".*" means "zero or more (*) of any character (.)".
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# 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".
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# Example 5:
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# Input:
# s = "mississippi"
# p = "mis*is*p*."
# Output: false
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#


class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        result = [[False for j in xrange(len(p) + 1)] for i in xrange(len(s) + 1)]
        
        result[0][0] = True
        for i in xrange(2, len(p) + 1):
            if p[i-1] == '*':
                result[0][i] = result[0][i-2]
                    
        for i in xrange(1,len(s) + 1):
            for j in xrange(1, len(p) + 1):
                if p[j-1] != '*':
                    result[i][j] = result[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.')
                else:
                    result[i][j] = result[i][j-2] or (result[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))
                    
        return result[len(s)][len(p)]