解法一

• 思路：
  写的第一个版本，知道是动态规划，但是不够简洁，因为动态方程 根本就没有写明白！！！！有点暴力的意思，其中还用到了剪枝操作
• 代码：

class Solution:
    # @param {string} s
    # @param {string} p
    # @return {boolean}
    def __init__(self):
        self.isRepeat = []
    def isMatch(self, s, p):
        print 's=', s, 'p=', p

        #剪枝操作
        tmp  = [s, p]
        if tmp in self.isRepeat:
            return
        else:
            self.isRepeat.append(tmp)

        #when len(s) == 0
        if len(s) == 0:
            if (len(p) == 2 and p[1] == '*') or len(p) == 0:
                return True
            if len(p) >= 2 and p[1] == '*':
                return self.starMatch(s, p)
            else:
                return False
        if len(s) != 0 and len(p) == 0:
            return False
        #when when len(p) == 1,which means that the process would go without 'a*' or '.*'
        if len(p) == 1:
            if p[0] == '*':
                return False
            elif p[0] == '.' and len(s) == 1:
                return True
            elif s == p:
                return True
            else:
                return False
        #p[0] == * represents no regular matches
        if p[0] == '*':
            return False
        if p[1] != '*':
            if p[0] == '.':
                return self.isMatch(s[1:], p[1:])
            else:
                if p[0] == s[0]:
                    return self.isMatch(s[1:], p[1:])
                else:
                    return False
        else:
            return self.starMatch(s, p)

    def starMatch(self, s, p):
        ''' when p[1] == '*', we need to discuss whether p[0] matches or not '''
        if len(s) != 0 and (p[0] == s[0] or p[0] == '.'):
            return  self.isMatch(s[1:], p) or self.isMatch(s, p[2:])
        else:
            return self.isMatch(s, p[2:])


def main():
    test = Solution()
    print test.isMatch("aaaaaaaaaaaaab", "a*a*a*a*a*a*a*a*a*a*c")

if __name__ == '__main__':
    main()

• AC
  但是时间上比较不理想

解法二

• 思路：

• 代码：

''' Title: Regular Expression Matching Url: https://leetcode.com/problems/regular-expression-matching/ Author: halibut735 Data: 15.6.24 '''
import pdb
class Solution:
    # @param {string} s
    # @param {string} p
    # @return {boolean}
    def __init__(self):
        self.count = 0
    def isMatch(self, s, p):
        self.whileloop = 0
        self.count += 1
        print 's=', s, 'p=', p
        pdb.set_trace()
        ''' boundary conditions... '''
        #when len(s) == 0
        if len(s) == 0 and len(p) == 0:
            return True
        if len(s) != 0 and len(p) == 0:
            return False
        #when when len(p) == 1,which means that the process would go without 'a*' or '.*'
        #p[0] == * represents no regular matches
        if p[0] == '*':
            return False


        #when p[1] != * or len(p) == 1
        if len(p) == 1 or p[1] != '*':
            print 'if count' , self.count
            if p[0] == '.' or (len(s) != 0 and p[0] == s[0]):
                return self.isMatch(s[1:], p[1:])
            else:
                return False
        else:
            print 'else count', self.count
            return self.starMatch(s, p)

    def starMatch(self, s, p):
        #when p[1] == '*', we need to discuss whether p[0] matches or not
        while p[0] == '.' or (len(s) != 0 and p[0] == s[0]):
            self.whileloop += 1
            print 'whileloop: ', self.whileloop
            if self.isMatch(s[1:], p):
                return True
        return self.isMatch(s, p[2:])


def main():
    test = Solution()
    print test.isMatch('a', "a*a")

if __name__ == '__main__':
    main()

• result: 进入了endless loop，还没找出原因，应该比较简单。。回头再找吧！！！别忘了

Created with Raphaël 2.1.2 Memo 6.24 me 6.24 me halibut735 halibut735 记得找bug OK, maybe before 7.1 痛苦ing… Created with Raphaël 2.1.2

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