？[4]621. Task Scheduler(Java)

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本文介绍了一种解决CPU任务调度问题的算法，旨在找到完成一系列字符型任务所需的最短时间间隔，考虑了冷却间隔n的限制条件。通过两种不同的实现方式探讨了如何优化任务调度过程。

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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.

Note:

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

代码1：

class Solution {
    public int leastInterval(char[] tasks, int n) {
        int[] alpNum = new int[26];
        for (int i = 0; i < tasks.length; i ++) {
            alpNum[tasks[i] - 'A'] ++;
        }
        Arrays.sort(alpNum);
        int i, maxNum = 0;
        for (i = 24; alpNum[i] == alpNum[25]; i --) {
            maxNum ++; 
        }
        return Math.max((alpNum[25] - 1) * n + alpNum[25] + maxNum, tasks.length);
    }
}

代码2：用优先队列实现（未懂？？？）

public int leastInterval(char[] tasks, int n) {
     Map<Character, Integer> map = new HashMap<>();
    for (int i = 0; i < tasks.length; i++) {
        map.put(tasks[i], map.getOrDefault(tasks[i], 0) + 1); // map key is TaskName, and value is number of times to be executed.
    }
    PriorityQueue<Map.Entry<Character, Integer>> q = new PriorityQueue<>( //frequency sort
            (a,b) -> a.getValue() != b.getValue() ? b.getValue() - a.getValue() : a.getKey() - b.getKey());

    q.addAll(map.entrySet());

    int count = 0;
    while (!q.isEmpty()) {
        int k = n + 1;
        List<Map.Entry> tempList = new ArrayList<>();
        while (k > 0 && !q.isEmpty()) {
            Map.Entry<Character, Integer> top = q.poll(); // most frequency task
            top.setValue(top.getValue() - 1); // decrease frequency, meaning it got executed
            tempList.add(top); // collect task to add back to queue
            k--;
            count++; //successfully executed task
        }

        for (Map.Entry<Character, Integer> e : tempList) {
            if (e.getValue() > 0) q.add(e); // add valid tasks 
        }

        if (q.isEmpty()) break;
        count = count + k; // if k > 0, then it means we need to be idle
    }
    return count;
}