Substring with Concatenation of All Words

题目：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).

思路：

定义两个hash表，一个就是传统的把所有的输入进去，对应代号。

本题的一个技巧就是每次从一个位置开始判断，然后再最外面写一个从当前位置判断再这个 start  能否成功的函数：

函数的技巧就是从start位置截取每一个单词的长度，比对是否出现过，出现的次数不能超过1次。

代码：

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> res;
        if(words.empty())   return res;
        int wordSize=words[0].size();
        int totalWords=words.size();
        int totalLen=wordSize*totalWords;
        
        if(s.size()<totalLen)   return res;
        
        unordered_map<string,int>wordCount;
        for(int i=0;i<totalWords;i++){
            wordCount[words[i]]++;
        }
        
        for(int i=0;i<=s.size()-totalLen;i++){
            if(checkSubstring( s,i, wordCount, wordSize, totalWords))   res.push_back(i);
        }
        return res;
    }
    
    bool checkSubstring(string &s,int start,unordered_map<string,int>&wordCount,int wordSize,int totalWords){
        if(s.size()-start+1<wordSize*totalWords)    return false;
        unordered_map<string,int>wordFound;
        
        for(int i=0;i<totalWords;i++){
            string tmp=s.substr(start+i*wordSize,wordSize);
            if(wordCount[tmp]==0)   return false;
            wordFound[tmp]++;
            if(wordFound[tmp]>wordCount[tmp])   return false;
        }
        
        return true;
    }
};