【LeetCode with Python】 Wildcard Matching

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原题页面：https://oj.leetcode.com/problems/wildcard-matching/
题目类型：动态规划
难度评价：★★★★★
本文地址：http://blog.csdn.net/nerv3x3/article/details/2921852

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.'*' Matches any sequence of characters (including the empty sequence).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", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

这道动态规划的题，其特殊test case造成的麻烦已经远远超过动态规划算法本身的复杂带来的麻烦了。之前在优化动态规划算法时，一直注重的是空间的优化，却没有考虑到有些cases是可以提前检测或进行预处理，从而降低运行时间的。如果不这样做，将一直卡在一些特别长的变态cases上Time Limit Exceeded。

1. 模式串p中连续的*可以预处理合并为一个*
2. 模式串中问号和字符的总数大于字符串s的长度时，匹配失败
3. 模式串中字符个数为0时，如果模式串中问号总数等于s长度，则匹配成功，否则匹配失败

class Solution:    # @param s, an input string    # @param p, a pattern string    # @return a boolean    def isMatch(self, s, p):        if (None == s or '' == s) and (None == p or '' == p):            return True        len_s = len(s)        len_p = len(p)        new_p = ''        last_ch = None        p_solid_char_num = 0        for i in range(0, len_p):            if '*' != last_ch or '*' != p[i]:                new_p += p[i]            last_ch = p[i]            if '*' != p[i]:                p_solid_char_num += 1        if p_solid_char_num > len_s:            return False        isAllStars = True        if (None == s or '' == s) and 0 == p_solid_char_num:                return True        p = new_p        len_p = len(new_p)        F = [[False for j in range(0, len_p + 1)] for i in range(0, 2)]        # 3 cases for True:        #    last ch matched, and this ch matcheds (right-down)        #    p comes *, and last matched (right)        #    last p.ch is *, s comes any ch, and last matched (down)        for i in range(1, len_s + 1):            if 1 == i:                F[0][0] = True                for j in range(1, len_p + 1):                    F[0][j] = (F[0][j - 1] and '*' == p[j - 1])            else:                for j in range(0, len_p + 1):          # cannot be F[0] = F[1]                    F[0][j] = F[1][j]            F[1][0] = False            for j in range(1, len_p + 1):                if True == F[0][j - 1] and (p[j - 1] == s[i - 1] or '?' == p[j - 1] or '*' == p[j - 1]):                    F[1][j] = True                elif (True == F[1][j - 1] or True == F[0][j]) and '*' == p[j - 1]:                    F[1][j] = True                else:                    F[1][j] = False        return F[1][len_p]