题意：

Given a task sequence tasks such as ABBABBC, and an integer k, which is the cool down time between two same tasks. Assume the execution for each individual task is 1 unit.

For example, if k = 3, then tasks BB takes a total of 5 unit time to finish, because B takes 1 unit of time to execute, then wait for 3 unit, and execute the second B again for another unit of time. So 1 + 3 + 1 = 5.

Given a task sequence and the cool down time, return the total execution time.

Follow up: Given a task sequence and the cool down time, rearrange the task sequence such that the execution time is minimal.

思路： 关注出现次数最多的那个字符串即可

代码：

unsigned min_runtime(vector<string>& vec, const int K) {
	if (vec.empty())  
		return 0; 
	unordered_map<string, int> hashtable;
	for (auto str : vec)
		hashtable[str]++;
	vector<int> cnt;
	for (auto elem : hashtable)
		cnt.push_back(elem.second);
	sort(cnt.rbegin(), cnt.rend());
	int frequency = 1;
	for (int i = 1; i < cnt.size(); ++i) {
		if (cnt[0] == cnt[i])
			frequency++;
		else
			break;
	}
	unsigned ret = cnt[0] + (cnt[0] - 1) * K + frequency - 1;
	return max(ret, (unsigned)vec.size()); 
}