leetcode: 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).

開兩個哈希表，一個記錄L中出現的詞，一個記錄搜索到的詞，如果搜索到的詞等於L中出現的詞，就推入下標

提交了好多遍，每個字符都可能是開頭，所以第一層循環是++i，這裏一開始饒了不少彎路

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        vector< int> res;
        const int word_length = L[0].size();
        const int words_length = word_length * L.size();
        if( S.size() == 0 || S.size() < words_length)
            return res;
        unordered_map< string, int> appear;
        unordered_map< string, int> expect;
        int appear_time = 0;
        for( int i = 0; i < L.size(); ++i)
            ++expect[L[i]];
        int tmpi;
        for( int i = 0; i < S.size() && S.size() - i >= words_length; ++i){
            string tmp = S.substr( i, word_length);
            if( expect[tmp] > 0){
                for( int j = i; j < S.size(); j = j + word_length){
                    tmp = S.substr( j, word_length);
                    if( expect[tmp] > 0 && appear[tmp] < expect[tmp]){
                        ++appear[tmp];
                        ++appear_time;
                    }
                    else{
                        appear.clear();
                        appear_time = 0;
                        break;
                    }
                    if( appear_time == L.size()){
                        res.push_back(i);
                        appear.clear();
                        appear_time = 0;
                        break;
                    }
                }
            }
        }
        return res;
    }
};