【LeetCode】30. Substring with Concatenation of All Words

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

题意：找出所给列表words中的单词相接的子句在s中出现的索引。

注意：words中的单词可能重复出现

思路：1.建立一个字典wdict，统计words中所有单词的个数

　　2.判断s中所有可能的子句是否符合要求

　　　　1）判断字据中每个单词是否出现在wdict中，没有的话，此子句不符合要求

　　　　2）子句符合要求的话，加入新的属于子句的字典，如果子句中某个单词出现的次数超过wdict中这个单词出现的个数，此子句不符合要求

 class Solution(object):

     def findSubstring(self, s, words):

         """

         :type s: str

         :type words: List[str]

         :rtype: List[int]

         """

         res = []

         length = len(words[0])

         num = len(words)

         l = length*num

         lens = len(s)

         wdict = {}

         sset = set()#建立集合，收集所有已经符合条件的字句，减少判断次数

         for word in words:

             wdict[word] = 1+(wdict[word] if word in wdict else 0)

         first_char = set(w[0] for w in words)#建立集合收集所有单词中的第一个字母，减少isRequired的次数

         for i in range(lens-l+1):

             if s[i] in first_char:

                 tmp = s[i:i+l]

                 if tmp in sset or self.isRequired(tmp,wdict,length):

                     sset.add(tmp)

                     res.append(i)

         return res

     def isRequired(self,s,wdict,length):

         comp = {}

         i = 0

         while(i<len(s)-length+1):

             tmp = s[i:i+length]

             if tmp not in wdict:

                 return False

             else:

                 comp[tmp] = 1+(comp[tmp] if tmp in comp else 0)

                 if comp[tmp]>wdict[tmp]:

                     return False

             i += length

         return True

巴特西

【LeetCode】30. Substring with Concatenation of All Words

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