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 wordsexactly 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).

 

 

下面这段代码只能通过三分之二的测试用例，因为a concatenation of each word in wordsexactly once，我能说题目意思给的不明确么？

网上还有应用“滑动窗口”的时间复杂度为O（n）算法，后面再分析！

1 class Solution { 2 public: 3     vector
     
       findSubstring(string s, vector
      
       & words) { 4         vector
       
         res; 5         int wordNum=words.size(),wordLen=words[0].size(); 6         if(s.size()
        
          dic,curWord; 9         for(int i=0;i
         
          dic[word])25                     break;26             }27             if(count==wordNum)28                 res.push_back(i);29         }30         return res;31     }32 };