Leetcode 621. Task Scheduler

最新推荐文章于 2022-05-16 22:52:29 发布

sundawei2016

最新推荐文章于 2022-05-16 22:52:29 发布

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本文介绍了一种解决CPU任务调度问题的算法，通过使用哈希表和优先队列，确保了在给定冷却间隔的情况下，CPU能以最少的时间完成所有任务。

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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.

Example 1:

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

我们要用一个hashmap 来储存每个字符和出现次数的映射关系，priority queue 来记录出现次数。

n = 2 是，每次都是 3个 3个地出现，所以cycle 是 3. 如果 pq 不为空，把 pq 最大的 n 个加入到pq 中，同时worktime 加一，再把每个 - 1 加回到pq 里。pq 还有书的话，则要加上一个 cycle，反之只用加上 worktime。

public int leastInterval(char[] tasks, int n) {        
        Map<Character, Integer> map = new HashMap<>();
        for (char t : tasks) map.put(t, map.getOrDefault(t, 0) + 1);
        Queue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        pq.addAll(map.values());
        int res = 0;
        int cycle = n + 1;
        while (!pq.isEmpty()) {
            int worktime = 0;
            List<Integer> tmp = new ArrayList<>();
            for (int i = 0; i < cycle; i++) {
                if (!pq.isEmpty()) {
                    tmp.add(pq.poll());
                    worktime++;
                }
            }
            for (int cnt : tmp) {
                if (--cnt > 0) pq.offer(cnt);
            }
            res += !pq.isEmpty() ? cycle : worktime;
        }
        return res;
    }