[LEETCODE][PYTHON][DP]REGULAR EXPRESSION MATCHING

# -*- coding: utf8 -*-‘‘‘https://oj.leetcode.com/problems/regular-expression-matching/

Implement regular expression matching with support for ‘.‘ and ‘*‘.

‘.‘ Matches any single character.‘*‘ Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:bool isMatch(const char *s, const char *p)

Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

===Comments by Dabay===这道题应该用动态规划的做法。我只想说，头大！

建立一个二维数组，第一维是s为基础，第二维是p为基础。（顺序很重要）初始化res[0]，长度比p多1，即多第一个true。同时考虑p中开始的几个偶数为是*的情况，初始化相应的true。其他为false。

然后每次放一个s中元素进来，尝试匹配整个p。

三种情况：（每次更新的是res[i+1][j+1]）    如果p[j]不是.和*，即是一般比较，同时观察res[i][j]。    如果p[j]是.，无需比较，根据res[i][j]的值即可。    如果p[j]是*，        考虑匹配零次的情况：判断res[i+1][j-1]是否为true        考虑匹配一次的情况：判断res[i+1][j]是否为true        考虑匹配多次的情况：判断字符是否相符，同时判断res[i][j+1]‘‘‘class Solution:    # @return a boolean    def isMatch(self, s, p):        res = []        defaultRow = [False] + [False for _ in p]        res.append(list(defaultRow))        res[0][0] = True        for x in xrange(len(p)):            if p[x] == ‘*‘:                res[0][x+1] = res[0][x-1]

        for i in xrange(len(s)):            res.append(list(defaultRow))            for j in xrange(len(p)):                if p[j] != ‘*‘:                    if (s[i] == p[j] or p[j] == ‘.‘) and res[i][j]:                        res[i+1][j+1] = True                elif p[j] == ‘.‘:                    if res[i][j]:                        res[i+1][j+1] = True                else: # *                    #匹配0次                    if res[i+1][j-1]:                        res[i+1][j+1] = True                    #匹配1次                    if res[i+1][j]:                        res[i+1][j+1] = True                    #匹配多次                    if (s[i] == p[j-1] or p[j-1] == ‘.‘) and res[i][j+1]:                        res[i+1][j+1] = True        return res[-1][-1]

def main():    sol = Solution()    s = "aab"    p = "c*a*b"    print sol.isMatch(s, p)

if __name__ == "__main__":    import time    start = time.clock()    main()    print "%s sec" % (time.clock() - start)