题解 | #字符串通配符#

def find(p, s):
    '''
    字符串通配符匹配判断
    :param p: 通配符
    :param s: 字符串
    :return: 匹配成功输出1,失败0
    '''
    lp, ls = len(p), len(s)
    '''
    初始化边界
    dp[0][0]=1, 空对空,匹配成功
    dp[0][j]=0, 空对非空,匹配失败
    dp[i][0]无法确定, 需要下边遍历确认
    '''
    # 构建动态规划列表 (ls+1)为列
    dp = [[0] * (ls + 1) for _ in range(lp + 1)]
    dp[0][0] = 1  # 初始化:空对空
    # 确定d[i][0]首列,一直是*就一直填1,碰到例外就终止
    for i in range(1, lp + 1):  # 第一列第二行开始遍历
        if p[i - 1] == '*':
            dp[i][0] = 1
        else:
            break
    # 确定其他格的值,遍历各行各列
    for i in range(1, lp + 1):
        for j in range(1, ls + 1):
            if p[i - 1] == '*' and (s[j - 1].isalnum() or dp[i-1][j]):
                dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
            elif p[i - 1] == '?' and s[j - 1].isalnum():
                dp[i][j] = dp[i - 1][j - 1]
            elif s[j - 1].lower() == p[i - 1].lower():
                dp[i][j] = dp[i - 1][j - 1]
    return dp[i][j]


if __name__ == '__main__':
    while 1:
        try:
            p, s = input(), input()
            if find(p, s):
                print('true')
            else:
                print('false')
        except:
            break