【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

 1 class Solution(object):
 2     def findSubstring(self, s, words):
 3         """
 4         :type s: str
 5         :type words: List[str]
 6         :rtype: List[int]
 7         """
 8         res = []
 9         length = len(words[0])
10         num = len(words)
11         l = length*num
12         lens = len(s)
13         wdict = {}
14         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
15         for word in words:
16             wdict[word] = 1+(wdict[word] if word in wdict else 0)
17         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
18         for i in range(lens-l+1):
19             if s[i] in first_char:
20                 tmp = s[i:i+l]
21                 if tmp in sset or self.isRequired(tmp,wdict,length):
22                     sset.add(tmp)
23                     res.append(i)
24         return res
25     def isRequired(self,s,wdict,length):
26         comp = {}
27         i = 0
28         while(i<len(s)-length+1):
29             tmp = s[i:i+length]
30             if tmp not in wdict:
31                 return False
32             else:
33                 comp[tmp] = 1+(comp[tmp] if tmp in comp else 0)
34                 if comp[tmp]>wdict[tmp]:
35                     return False
36             i += length
37         return True
38
39