LeetCode Substring with Concatenation of All Words, leetcode

LeetCode Substring with Concatenation of All Words, leetcode

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 shoshould return the indices:[0,9].
(Order does not matter ).

Locate the starting point of the substring that appears in all strings in string S, and the sequence of occurrence can be unordered.

Idea: it is a question for maintaining a dynamic window and maintaining a [start, end) window to contain strings in L, every moving unit is len (L strings are of the same size)

class Solution {public:    vector<int> findSubstring(string S, vector<string> &L) {        map<string, int> wordTimes;        for (int i = 0; i < L.size(); i++) {            if (wordTimes.count(L[i]) == 0)                 wordTimes[L[i]] = 1;            else wordTimes[L[i]]++;        }           int len = L[0].size();        vector<int> res;        for (int i = 0; i < len; i++) {            map<string, int> curTimes;            int start = i, cnt = 0;            for (int end = i; end <= (int)S.size()-len; end += len) {                string word = S.substr(end, len);                if (wordTimes.find(word) != wordTimes.end()) {                    if (curTimes.find(word) == curTimes.end())                        curTimes[word] = 1;                    else curTimes[word]++;                    if (curTimes[word] <= wordTimes[word])                        cnt++;                    else {                        for (int k = start; ; k += len) {                            string tmp = S.substr(k, len);                            curTimes[tmp]--;                            if (tmp == word) {                                start = k + len;                                break;                            }                            cnt--;                        }                    }                    if (cnt == L.size())                        res.push_back(start);                } else {                    start = end + len;                    curTimes.clear();                    cnt = 0;                }             }        }        return res;    }};