Task Scheduling Algorithms

最新推荐文章于 2024-06-25 00:00:00 发布

转载最新推荐文章于 2024-06-25 00:00:00 发布 · 141 阅读

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原文链接：http://www.cnblogs.com/codingforum/p/8180042.html

本文介绍了两种典型的任务调度问题：一种是在多个任务具有不同截止时间和完成时间的情况下，如何安排任务顺序以最小化超过截止时间的最大偏差；另一种是考虑到任务间的冷却间隔，如何安排任务以减少空闲周期并达到最短完成时间。

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Reference

[1] https://leetcode.com/articles/task-scheduler/

[2] https://www.hackerrank.com/challenges/task-scheduling/problem (hackerrank or interviewstreet)

[3] https://github.com/ningke/tasksched

[4] http://p--np.blogspot.co.uk/2011/07/segment-tree.html

[5] segment tree https://www.cnblogs.com/xiaoyao24256/p/6590885.html

[6] binary indexed tree https://www.cnblogs.com/whensean/p/6851018.html

You have a long list of tasks that you need to do today. Task i is specified by the deadline by which you have to complete it (Di) and the number of minutes it will take you to complete the task (Mi). You need not complete a task at a stretch. You can complete a part of it, switch to another task and then switch back.

You've realized that it might not actually be possible complete all the tasks by their deadline, so you have decided to complete them so that the maximum amount by which a task's completion time overshoots its deadline is minimized.

Solution: The key insight is that the task with a later deadline MUST be completed later than a task with an earlier deadline - regardless how you break up the tasks. Otherwise, the first task (the one with an earlier deadline) will overshoot its own deadline by more than the second task overshoots its deadline. So we can essentially just sort the tasks by their deadline and complete them in that order. The problem description about breaking up tasks are just there to confuse you.

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.

Note:

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

The tasks with the currently maximum number of outstanding (pending)instances will contribute to a large number of idle cycles in the future, if not executed with appropriate interleavings with the other tasks. Thus, we need to re-execute such a task as soon as its cooling time is finished.

Thus, based on the above ideas, firstly, we obtain a count of the number of instances of each task in

Now, the task picked up first after the sorting, will either be the first task picked up in the last iteration(which will now be picked after its cooling time has been finished) or the task picked will be the one which lies at

We stop this process, when the pending instances of all the tasks have been reduced to 0. At this moment,

public class Solution {
    public int leastInterval(char[] tasks, int n) {
        int[] map = new int[26];
        for (char c: tasks)
            map[c - 'A']++;
        Arrays.sort(map);
        int time = 0;
        while (map[25] > 0) {
            int i = 0;
            while (i <= n) {
                if (map[25] == 0)
                    break;
                if (i < 26 && map[25 - i] > 0)
                    map[25 - i]--;
                time++;
                i++;
            }
            Arrays.sort(map);
        }
        return time;
    }
}

转载于:https://www.cnblogs.com/codingforum/p/8180042.html