python写算法题：leetcode: 30. Substring with Concatenation of All Words

https://leetcode.com/problems/substring-with-concatenation-of-all-words/#/description

class Solution(object):    def matchstr(self, p0, p1, wordsind, wlen, posmap):        wv={}        for w in wordsind.keys():            wv[w]=0        for i in xrange(p0, p1, wlen):            if i in posmap:                wv[posmap[i]]+=1        return wv    def findSubstring(self, s, words):        """        :type s: str        :type words: List[str]        :rtype: List[int]        """        if len(words)==0: return []        wlen = len(words[0])        wordsind = {}        wcnt={}        posmap={}        for w in words:            if w in wordsind:                wcnt[w]+=1                continue            else:                matchlst=[]                wcnt[w]=1            i=0            while i>=0:                i = s.find(w, i)                if i>=0:                    posmap[i]=w                    matchlst.append(i)                    i+=1            wordsind[w]=matchlst        res=[]        endpos=len(words)*wlen-wlen        for p0 in xrange(len(words[0])):            wv = self.matchstr(p0, p0+len(words)*wlen, wordsind, wlen, posmap)            if wv == wcnt:                res.append(p0)            for p in xrange(p0+wlen, len(s)-endpos, wlen):                if p-wlen in posmap:                    wv[posmap[p-wlen]]-=1                if p+endpos in posmap:                    wv[posmap[p+endpos]]+=1                if wv == wcnt:                    res.append(p)        return res

思路：充分利用匹配串长度相同规则，开头匹配偏移取值是单个匹配窜的长度范围，然后跳跃式的检查是否满足要求，最终运算复杂度为 O(n+m)