Substring with Concatenation of All Words

/*

Substring with Concatenation of All Words

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).
--O(n) not optimized

*/
class Solution {
public:
    vector<string>::iterator findWord(vector<string> L, string input){
      vector<string>::iterator it;
      for(it=L.begin();it<L.end();it++){
        if(*it==input){
          return it;
        }
      }
      return L.end();
    }
    vector<int> findSubstring(string S, vector<string> &L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result;
        int numOfWord = L.size();
        if(numOfWord==0) return result;
        int wordSize = L.at(0).length();
        int pos = 0;
        int size = S.length();
        int currStart;
        vector<string>::iterator found;
        vector<string> temp;
  bool wordExist = false;
        while(pos<size-numOfWord*wordSize+1){
          temp.clear();
          temp.insert(temp.begin(),L.begin(),L.end());
          currStart = pos;
          while(temp.size()!=0 && currStart<size-wordSize+1){
            //found = findWord(L, S.substr(currStart,wordSize));
   wordExist = false;
   for(found=temp.begin();found<temp.end();found++){
    if(found!=temp.end()&&*found==S.substr(currStart,wordSize)){
     temp.erase(found);
     wordExist = true;
     break;
    }
   }
            if(wordExist){
            currStart = currStart+wordSize;
            }
   else break;
          }
          if(temp.size()==0){
           result.push_back(pos);
          }
          pos++;
        }
        return result;
    }
};