题解 | #字符串通配符#

while True:
    try:
        s1 = input()
        s2 = input()
        # 初始化dp数组
        dp = [[False for j in range(len(s2) + 1)] for i in range(len(s1) + 1)]
        dp[0][0] = True
        # 第0行全部为False(因为空字符无法匹配非空字符)
        for j in range(1, len(s2) + 1):
            dp[0][j] = False
        # 处理第0列，因为这一列可以提前预判
        for i in range(1, len(s1) + 1):
            if s1[i - 1] == "*" or s1[i - 1] == "?":  # 当模式为*或者?时，能匹配
                dp[i][0] = True
            else:
                break  # why?
        for i in range(1, len(s1) + 1):
            for j in range(1, len(s2) + 1):
                if s1[i - 1].lower() == s2[j - 1].lower():
                    dp[i][j] = dp[i - 1][j - 1]
                elif s1[i - 1] == "?" and (
                    (s2[j - 1].isalnum()) or (s2[j - 1].isalpha())
                ):  # s2[j-1]不为特殊符号
                    dp[i][j] = dp[i - 1][j - 1]
                elif s1[i - 1] == "*":
                    dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
        if dp[len(s1)][len(s2)]:
            print("true")
        else:
            print("false")
    except:
        break