class Solution(object):# p为匹配模式，s为字符串def recursive(self, s, p, si, pi, cur):first = Truen_cur = curwhile si < len(s) and pi < len(p) and (s[si] == p[pi] or p[pi] == '?'):si += 1pi += 1if pi == len(p):return si == len(s)if p[pi] == '*':while pi < len(p) and p[pi] == '*':pi += 1if pi >= len(p):return Truefor i in range(si, len(s)):# 表明开始重合，从这里再度开始递归if p[pi] != s[i] and p[pi] != '?':continueif first:cur += 1first = False# 可能存在多次重合但是还不算真正匹配的情况if self.recursive(s, p, i, pi, cur + 1):return Trueif cur > n_cur + 1: # 正常来说n_cur = cur + 1return Falsereturn Falsedef isMatch(self, s, p):"""
        :type s: str
        :type p: str
        :rtype: bool
        """return self.recursive(s, p, 0, 0, 0)

class Solution(object):# p为匹配模式，s为字符串def isMatch(self, s, p):"""
        :type s: str
        :type p: str
        :rtype: bool
        """if len(s) != len(p) - p.count('*'):return Falsenewp = ""i = 0while i < len(p):newp += p[i]if p[i] == '*':while i + 1 < len(p) and p[i + 1] == '*':i += 1i += 1sl, pl = len(s), len(newp)dp = [[False for x in range(pl + 1)] for y in range(sl + 1)]dp[0][0] = Trueif pl > 0 and p[0] == '*':dp[0][1] = Truefor x in range(1, sl + 1):for y in range(1, pl + 1):if newp[y - 1] != '*':dp[x][y] = dp[x - 1][y - 1] and (s[x - 1] == newp[y - 1] or newp[y - 1] == '?')else:dp[x][y] = dp[x - 1][y] or dp[x][y - 1]return dp[sl][pl]

class Solution:# @return a booleandef isMatch(self, s, p):length = len(s)if len(p) - p.count('*') > length:return Falsedp = [True] + [False]*lengthfor i in p:if i != '*':# 因为依赖项是前面的值，所以不能从前面往后面扫，得从后往前计算for n in reversed(range(length)):dp[n+1] = dp[n] and (i == s[n] or i == '?')else:# 更新当前行的数据for n in range(1, length+1):dp[n] = dp[n-1] or dp[n]dp[0] = dp[0] and i == '*'return dp[-1]

class Solution(object):# p为匹配模式，s为字符串def isMatch(self, s, p):si, pi = 0, 0lastmatch, laststar = -1, -1sl, pl = len(s), len(p)if pl - p.count('*') > sl:return False# 注意条件顺序while si < sl:if pi < pl and (s[si] == p[pi] or p[pi] == '?'):pi += 1si += 1elif pi < pl and p[pi] == '*':lastmatch, laststar = si, pi  # 之所以不更新lastmatch是因为考虑到*只匹配0个字符串pi += 1# 再次进到这个判断，说明当前下标对应的值不相等elif laststar != -1:pi = laststar + 1  # pi当前不是*，并且回到上一次星的后面，专门用来给si匹配lastmatch += 1  # 必须更新lastmatch，因为之前已经不想等，如果在回到开始的状态就会陷入死循环si = lastmatchelse:return False# 可能存在p的末尾都是*的情况while pi < len(p) and p[pi] == '*':pi += 1# 最后匹配成功模式字符串的下标必然为其长度，表示已经匹配完成return pi == pl

class Solution(object):def isMatch(self, s, p):"""
        :type s: str
        :type p: str
        :rtype: bool
        """seen = {}wild_single, wild_multi = "?", "*"# seen has the pattern - source tuple as key, and bool result as successsource, pattern = s, pdef is_match(sindex, pindex):key = (sindex, pindex)if key in seen:return seen[key]result = True# if there's no string, and pattern is not only * then failif sindex >= len(source):for wildindex in xrange(pindex, len(pattern)):if pattern[wildindex] != wild_multi:result = Falsebreak# there's a string, but no patternelif pindex >= len(pattern):result = False# if next pattern is multi though, that's somethingelif pattern[pindex] == wild_multi:# for zero, simply check sindex, pindex + 1result = is_match(sindex, pindex + 1) # just for easier debug# if zero, than it's a match# otherwise we need to check multi# for that, if char is not a wild, then it has to match the source,result = result or is_match(sindex + 1, pindex)else:# either a regular char, or wild_singleresult = (( pattern[pindex] == wild_single or pattern[pindex] == source[sindex]) and is_match(sindex + 1, pindex + 1))seen[key] = resultreturn resultif (len(p) - p.count(wild_multi) > len(s)):return Falsereturn is_match(0, 0)