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 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).

Analyse: Look at the code below. 

Runtime: 1548ms. 

class Solution {
public:
    //first scan the words, and put each word and its occurance time into an map
    //then gradually scan the the string, create a new map and compare it with original map
    //if equal, push the first index into result vector
    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> result;
        if(s.empty() || words.empty()) return result;
        int wordsNumber = words.size();
        int wordLen = words[0].size();
        int allWordsLen = wordsNumber * wordLen;
        if(s.size() < allWordsLen) return result;
        
        unordered_map<string, int> um;
        for(int i = 0; i < words.size(); i++)
            um[words[i]]++;
            
        for(int i = 0; i < s.size() - allWordsLen + 1; i++){
            unordered_map<string, int> tempUm;
            for(int j = 0; j < wordsNumber; j++){
                string temp = s.substr(i + j * wordLen, wordLen);
                tempUm[temp]++;
            }
            if(tempUm == um)
                result.push_back(i);
        }
        return result;
    }
};

After adding a judgement, the run time has been reduced to 596ms.

class Solution {
public:
    //first scan the words, and put each word and its occurance time into an map
    //then gradually scan the the string, create a new map and compare it with original map
    //if equal, push the first index into result vector
    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> result;
        if(s.empty() || words.empty()) return result;
        int wordsNumber = words.size();
        int wordLen = words[0].size();
        int allWordsLen = wordsNumber * wordLen;
        if(s.size() < allWordsLen) return result;
        
        unordered_map<string, int> um;
        for(int i = 0; i < words.size(); i++)
            um[words[i]]++;
            
        for(int i = 0; i < s.size() - allWordsLen + 1; i++){
            //find the first matched word
            bool found = false;
21             for(int k = 0; k < wordsNumber; k++){
22                 if(s.substr(i, wordLen) == words[k]){
23                     found = true;
24                     break;
25                 }
26             }
27             if(!found) continue;
            
            unordered_map<string, int> tempUm;
            for(int j = 0; j < wordsNumber; j++){
                string temp = s.substr(i + j * wordLen, wordLen);
                tempUm[temp]++;
            }
            if(tempUm == um)
                result.push_back(i);
        }
        return result;
    }
};