原题地址：https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/

题意：

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

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

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

解题思路：使用一个字典统计一下L中每个单词的数量。由于每个单词的长度一样，以题中给的例子而言，可以3个字母3个字母的检查，如果不在字典中，则break出循环。有一个技巧是建立一个临时字典curr，用来统计S中那些在L中的单词的数量，必须和L中单词的数量相等，否则同样break。

代码：

class Solution:
    # @param S, a string
    # @param L, a list of string
    # @return a list of integer
    def findSubstring(self, S, L):
        words={}
        wordNum=len(L)
        for i in L:
            if i not in words:
                words[i]=1
            else:
                words[i]+=1
        wordLen=len(L[0])
        res=[]
        for i in range(len(S)+1-wordLen*wordNum):
            curr={}; j=0
            while j<wordNum:
                word=S[i+j*wordLen:i+j*wordLen+wordLen]
                if word not in words: 
                    break
                if word not in curr: 
                    curr[word]=1
                else:
                    curr[word]+=1
                if curr[word]>words[word]: break
                j+=1
            if j==wordNum: res.append(i)
        return res