原题传送门

1. 题目描述

2. Solution 1: 转换为正则匹配

def solve1(s, p):
    p = p.lower()
    s = s.lower()
    p = p.replace('.', '\.').replace('?', '[0-9a-z]').replace('*', '#')
    p = re.sub('#+', '[0-9a-z]*', p)
    if bool(re.fullmatch(p, s)):
        print('true')
    else:
        print('false')

3. Solution 2: DP

import re
import sys

if sys.platform != "linux":
    sys.stdin = open("input/HJ71.txt")


def solve1(s, p):
    p = p.lower()
    s = s.lower()
    p = p.replace('.', '\.').replace('?', '[0-9a-z]').replace('*', '#')
    p = re.sub('#+', '[0-9a-z]*', p)
    if bool(re.fullmatch(p, s)):
        print('true')
    else:
        print('false')


def solve2(s, p):
    m, n = len(p), len(s)
    '''
    初始化边界
    1. dp[0][0] = True, 空模式空字符，匹配成功
    2. dp[0][j] = False, 空模式无法匹配非空字符串
    3. dp[i][0] 不确定，只有模式为*时，才能匹配空字符串，即模式p前i个字符均为* 
    '''
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = True
    for i in range(1, m + 1):
        if p[i - 1] != '*':
            break
        dp[i][0] = True

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if p[i - 1] == '*':
                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]

    if dp[m][n]:
        print("true")
    else:
        print("false")


while True:
    try:
        p = input().strip()
        s = input().strip()
        solve2(s, p)
    except:
        break

标签：lower,re,replace,牛客,range,HJ71,print,机试,dp
来源： https://www.cnblogs.com/junstat/p/16176135.html