Leetcode（python）--10.Regular Expression Matching

如：

isMatch(“aa”,”a”) → false
isMatch(“aa”,”aa”) → true
isMatch(“aaa”,”aa”) → false
isMatch(“aa”, “a*”) → true
isMatch(“aa”, “.*”) → true
isMatch(“ab”, “.*”) → true
isMatch(“aab”, “c*a*b”) → true

不同于我们平时用的正则表达式，这道题限制成了简易正则表达式，只有两个规则。

python代码解决如下（ps：借鉴了网上的答案）

class Solution(object):    def isMatch(self,s, p):  # s能被p匹配        """         :type s: str         :type p: str         :rtype: bool        """        lens = len(s)        lenp = len(p)        res = []        for i in range(lens + 1):            t = []            for j in range(lenp + 1):                t.append(False)            res.append(t)        res[lens][lenp] = True        # res = [[False for i in range(lenp + 1)] for i in range(lens + 1)]        for i in range(lenp - 1, -1, -1):  # lenp-1到-1(不包含-1即到0)            if p[i] == "*":  # 为什么逆着，因为*号总在右边                res[lens][i] = res[lens][i + 1]            elif i + 1 < lenp and p[i + 1] == "*":                res[lens][i] = res[lens][i + 1]            else:                res[lens][i] = False        for i in range(lens - 1, -1, -1):  # s的所有下标从大到小            for j in range(lenp - 1, -1, -1):  # p的所有下标                if p[j] == "*":                    if j - 1 >= 0 and p[j - 1] != "*":  # 注意两个*不能匹配                        res[i][j] = res[i][j + 1]                    else:                        return False                elif j + 1 < lenp and p[j + 1] == "*":                    if s[i] == p[j] or p[j] == ".":                        res[i][j] = res[i][j + 2] or res[i + 1][j] or res[i + 1][j + 2]                    else:                        res[i][j] = res[i][j + 2]                else:                    if s[i] == p[j] or p[j] == ".":                        res[i][j] = res[i + 1][j + 1]                    else:                        res[i][j] = False        return res[0][0]