题解 | #字符串通配符#

# 2024年9月24日
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
        # 优先处理第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] == '?':    # 
                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():
                    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