leetcode 621. Task Scheduler

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本文探讨了CPU在处理多个任务时如何有效安排任务执行顺序的问题。特别地，考虑到两个相同任务之间必须间隔一定数量的不同任务或者空闲周期，文章提供了一种算法来计算完成所有任务所需的最短时间，并附带Python实现。

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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:

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

这题主要恶心在分析上，公式是（k-1）*(n+1)+p,k代表频率最高的那个值，p代表频率最高元素的个数。。。然后特殊情况是计算得到的结果小于task的个数，此时答案就是task的个数。

class Solution(object):
    def leastInterval(self, tasks, n):
        """
        :type tasks: List[str]
        :type n: int
        :rtype: int
        """
        counts=collections.Counter(tasks)
        maximal=max(counts.items(),key=lambda x:x[1])
        p=0
        for key in counts:
            if counts[key]==maximal[1]:
                p+=1
        tmp=(maximal[1]-1)*(n+1)+p
        if tmp<len(tasks):
            return len(tasks)
        return tmp