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).
题意：找出S串中出现L所有串的子串的起始点，，出现的次序可以无序。
思路：是一道维护一个动态窗口的题目，维护一个[start, end)的窗口来包含L中的串，每次移动的单位都是len(L的串都是同样大小的)
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;}};
哪里会顾得上这些。等到时间将矛盾一层层降解为流言是非误解过结
