44. Wildcard Matching

# Time:  O(m + n) ~ O(m * n)
# Space: O(1)

# iterative solution with greedy
class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        count = 0  # used for complexity check
        p_ptr, s_ptr, last_s_ptr, last_p_ptr = 0, 0, -1, -1
        while s_ptr < len(s):
            if p_ptr < len(p) and (s[s_ptr] == p[p_ptr] or p[p_ptr] == '?'):
                s_ptr += 1
                p_ptr += 1
            elif p_ptr < len(p) and p[p_ptr] == '*':
                p_ptr += 1
                last_s_ptr = s_ptr
                last_p_ptr = p_ptr
            elif last_p_ptr != -1:
                last_s_ptr += 1
                s_ptr = last_s_ptr
                p_ptr = last_p_ptr
            else:
                assert(count <= (len(p)+1) * (len(s)+1))
                return False
            count += 1  # used for complexity check
 
        while p_ptr < len(p) and p[p_ptr] == '*':
            p_ptr += 1
            count += 1  # used for complexity check

        assert(count <= (len(p)+1) * (len(s)+1))
        return p_ptr == len(p)


# dp with rolling window
# Time:  O(m * n)
# Space: O(n)
class Solution2(object):
    # @return a boolean
    def isMatch(self, s, p):
        k = 2
        result = [[False for j in xrange(len(p) + 1)] for i in xrange(k)]

        result[0][0] = True
        for i in xrange(1, len(p) + 1):
            if p[i-1] == '*':
                result[0][i] = result[0][i-1]
        for i in xrange(1,len(s) + 1):
            result[i % k][0] = False
            for j in xrange(1, len(p) + 1):
                if p[j-1] != '*':
                    result[i % k][j] = result[(i-1) % k][j-1] and (s[i-1] == p[j-1] or p[j-1] == '?')
                else:
                    result[i % k][j] = result[i % k][j-1] or result[(i-1) % k][j]

        return result[len(s) % k][len(p)]


# dp
# Time:  O(m * n)
# Space: O(m * n)
class Solution3(object):
    # @return a boolean
    def isMatch(self, s, p):
        result = [[False for j in xrange(len(p) + 1)] for i in xrange(len(s) + 1)]

        result[0][0] = True
        for i in xrange(1, len(p) + 1):
            if p[i-1] == '*':
                result[0][i] = result[0][i-1]
        for i in xrange(1,len(s) + 1):
            result[i][0] = False
            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-1] or result[i-1][j]

        return result[len(s)][len(p)]


# recursive, slowest, TLE
class Solution4(object):
    # @return a boolean
    def isMatch(self, s, p):
        if not p or not s:
            return not s and not p

        if p[0] != '*':
            if p[0] == s[0] or p[0] == '?':
                return self.isMatch(s[1:], p[1:])
            else:
                return False
        else:
            while len(s) > 0:
                if self.isMatch(s, p[1:]):
                    return True
                s = s[1:]
            return self.isMatch(s, p[1:])

Solution from kamyu104/LeetCode-Solutions · MIT