[leetcode] 621. Task Scheduler @ python

原题

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.

Note:

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

解法

参考: 【LeetCode】621. Task Scheduler 解题报告（Python & C++）
贪心算法, 我们首先排列频率最高的字母, 如果几个字母频率都是最高, 那么把它们当做一体. 例如AAAABBBBCCDE，n=3。那么我们将具有相同个数的任务A和B视为一体，最终满足要求的分配为：AB - - AB - - AB - - AB

剩余的任务在不违背要求间隔的情况下穿插进间隔位置即可，在本例中则是ABCDABCEAB - - AB. 假设频率小的字母数量很少, 那么结果=（最大频数-1）*（n + 1） + （最大频数的字母个数）。假设频率小的字母数量很多, 那么将频率小的字母穿插进来(频率大于1时)或者直接加到末尾(频率等于1时)即可, 结果取max(res, len(tasks))

代码

class Solution(object):
    def leastInterval(self, tasks, n):
        """
        :type tasks: List[str]
        :type n: int
        :rtype: int
        """
        count = collections.Counter(tasks)
        max_freq = count.most_common()[0][1]
        num_of_max_freq = len([k for k, v in count.items() if v == max_freq])
        # if we ignore the rest letters
        res = (n+1)*(max_freq - 1) + num_of_max_freq
        # if the rest letters are very long
        return max(res, len(tasks))