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Lintcode 945. Task Scheduler (Medium) (Python)

程序员文章站 2022-07-15 12:15:47
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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.
Example

Given tasks = [‘A’,’A’,’A’,’B’,’B’,’B’], n = 2, return 8.

Explanation:
A -> B -> idle -> A -> B -> idle -> A -> B.

Notice

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

Code:

1.

class Solution:
    """
    @param tasks: the given char array representing tasks CPU need to do
    @param n: the non-negative cooling interval
    @return: the least number of intervals the CPU will take to finish all the given tasks
    """
    def leastInterval(self, tasks, n):
        # write your code here
        cnt=collections.Counter(tasks)
        vmax = max(cnt.values())
        count = 0
        for i in cnt:
            if cnt[i]==vmax:
                count+=1
        return max(len(tasks), (vmax-1)*(n+1)+count)

2.

class Solution:
    """
    @param tasks: the given char array representing tasks CPU need to do
    @param n: the non-negative cooling interval
    @return: the least number of intervals the CPU will take to finish all the given tasks
    """
    def leastInterval(self, tasks, n):
        # write your code here
        cnt = collections.Counter(tasks)
        vmax = max(cnt.values())
        slots = (vmax - 1) * n
        return max(len(tasks), len(tasks)+slots+vmax-1-sum(n-(n==vmax) for n in cnt.values()))
相关标签: Lintcode Python