【LeetCode】621. Task Scheduler

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621. Task Scheduler

Description：
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks. Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.
However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.
You need to return the least number of intervals the CPU will take to finish all the given tasks.
Difficulty：Medium
Example:

Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.

Note:
The number of tasks is in the range [1, 10000].
The integer n is in the range [0, 100].

方法1：数学推导

• Time complexity : $O\left ( n \right )$
• Space complexity : $O\left ( n \right )$
  思路：
  先找到出现次数最多的元素，然后试此元素的中间间隔为n；
  然后将其他元素插空；
  最后会发现有可能会有元素只能在最后，取个max即可。

class Solution {
public:
	int leastInterval(vector<char>& tasks, int n) {
		unordered_map<char, int> hash_map;
		int count = 0;
		for (auto t : tasks) {
			hash_map[t]++;
			count = max(hash_map[t], count);
		}
		int res = (count - 1) * (n + 1);
		for (auto t : hash_map)
			if (t.second == count)
				res++;
		return max((int)tasks.size(), res);
    }
};