LeetCode Substring with Concatenation of All Words

问题网址：https://leetcode.com/problems/substring-with-concatenation-of-all-words/description/

问题描述：
给你一个字符串s和一个长度相同的单词列表。 找出s中所有子字符串的起始索引，这些字符串中的每个单词只是一个字符串，没有任何中介字符。

例如，给出：
s：“barfoothefoobarman”
词语：[“foo”，“bar”]

你应该返回索引：[0,9]。
（顺序无关紧要）。

一个O(N)的解法

 // travel all the words combinations to maintain a window
    // there are wl(word len) times travel
    // each time, n/wl words, mostly 2 times travel for each word
    // one left side of the window, the other right side of the window
    // so, time complexity O(wl * 2 * N/wl) = O(2N)
    vector<int> findSubstring(string S, vector<string> &L) {
        vector<int> ans;
        int n = S.size(), cnt = L.size();
        if (n <= 0 || cnt <= 0) return ans;

        // init word occurence
        unordered_map<string, int> dict;
        for (int i = 0; i < cnt; ++i) dict[L[i]]++;

        // travel all sub string combinations
        int wl = L[0].size();
        for (int i = 0; i < wl; ++i) {
            int left = i, count = 0;
            unordered_map<string, int> tdict;
            for (int j = i; j <= n - wl; j += wl) {
                string str = S.substr(j, wl);
                // a valid word, accumulate results
                if (dict.count(str)) {
                    tdict[str]++;
                    if (tdict[str] <= dict[str]) 
                        count++;
                    else {
                        // a more word, advance the window left side possiablly
                        while (tdict[str] > dict[str]) {
                            string str1 = S.substr(left, wl);
                            tdict[str1]--;
                            if (tdict[str1] < dict[str1]) count--;
                            left += wl;
                        }
                    }
                    // come to a result
                    if (count == cnt) {
                        ans.push_back(left);
                        // advance one word
                        tdict[S.substr(left, wl)]--;
                        count--;
                        left += wl;
                    }
                }
                // not a valid word, reset all vars
                else {
                    tdict.clear();
                    count = 0;
                    left = j + wl;
                }
            }
        }

        return ans;
    }