【leetcode】44. Wildcard Matching

题目如下：

解题思路：本题和【leetcode】97. Interleaving String非常相似，同样可以采用动态规划的方法。记dp[i][j] = 1或者0 表示pattern[0:i]是否匹配string[0:j] ，如果pattern[i] == string[j] 或者 pattern[i] == ‘?‘,那么dp[i][j]  = dp[i-1][j-1]；如果pattern[i] = ‘*‘ 则复杂一些，因为‘*‘可以匹配s[j-1]，s[j] 或者不匹配，因此只要满足其中任意一种情况都视为匹配，得出递推关系式：dp[i][j] = max(dp[i-1][j-1],dp[i][j-1],dp[i-1][j])。

代码如下：

class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        np = ‘‘

        #合并连续出现的*,提高效率
        for i in p:
            if len(np) == 0:
                np += i
            elif np[-1] == i and i == ‘*‘:
                continue
            else:
                np += i

        #pattern和string都加上‘#‘开头，处理pattern以*开头，但是又不做匹配的情况
        np  = ‘#‘ + np
        p = np
        s = ‘#‘ + s

        dp = [[0] * len(s) for i in p]
        #print dp

        dp[0][0] = 1

        for i in range(len(dp)):
            for j in range(len(dp[i])):
                if p[i] == ‘*‘:
                    if i > 0 and j > 0:
                        dp[i][j] = max(dp[i-1][j-1],dp[i][j-1],dp[i-1][j])
                    elif i > 0 and j == 0:
                        dp[i][j] = dp[i-1][j]
                    elif i == 0 and j > 0:
                        dp[i][j] = dp[i][j-1]
                elif i > 0 and j > 0 and (p[i] == ‘?‘ or  p[i] == s[j]):
                    dp[i][j] = dp[i-1][j-1]
        #print dp
        return dp[-1][-1] == 1

原文地址：https://www.cnblogs.com/seyjs/p/9559148.html