[HashTable]030|[HashTable]030 Substring with Concatenation of All Words [HashTable]030SubstringwithConcatena

分类：HashTable
考察知识点：HashTable 数组遍历
最优解时间复杂度：O(nm+n)*

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.
Example 1:

Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9] Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively. The output order does not matter, returning [9,0] is fine too.

Example 2:

Input: s = "wordgoodstudentgoodword", words = ["word","student"] Output: []

代码：
解法:

class Solution: def findSubstring(self, s, words): """ :type s: str :type words: List[str] :rtype: List[int] """ #先判定边界条件 if len(s)==0 or len(words)==0: return []#先把这个words存到 words_length=len(words) word_length=len(words[0])if len(s)0): if s[start:start+word_length] not in words_dict_copy: i break if words_dict_copy[s[start:start+word_length]]==0: break words_dict_copy[s[start:start+word_length]]-=1 start+=word_length count-=1 if count==0: res.append(i) return res

讨论： 【[HashTable]030|[HashTable]030 Substring with Concatenation of All Words】1.这道题目是道Hard题，我就不深究它了，Beats的人少就少吧
2.这道题的思路是首先判断是否可以返回空的几个条件，然后再是把words全部放到dict中去，然后在每次循环的时候弄一个dict_copy,再循环判断dict_copy中的数据