给你一个用字符数组 tasks 表示的 CPU 需要执行的任务列表。其中每个字母表示一种不同种类的任务。任务可以以任意顺序执行，并且每个任务都可以在 1 个单位时间内执行完。在任何一个单位时间，CPU 可以完成一个任务，或者处于待命状态。
然而，两个 相同种类 的任务之间必须有长度为整数 n 的冷却时间，因此至少有连续 n 个单位时间内 CPU 在执行不同的任务，或者在待命状态。
你需要计算完成所有任务所需要的 最短时间 。
示例 1：
输入：tasks = ["A","A","A","B","B","B"], n = 2
输出：8
解释：A -> B -> (待命) -> A -> B -> (待命) -> A -> B
在本示例中，两个相同类型任务之间必须间隔长度为 n = 2 的冷却时间，而执行一个任务只需要一个单位时间，所以中间出现了（待命）状态。
示例 2：
输入：tasks = ["A","A","A","B","B","B"], n = 0 输出：6 解释：在这种情况下，任何大小为 6 的排列都可以满足要求，因为 n = 0 ["A","A","A","B","B","B"] ["A","B","A","B","A","B"] ["B","B","B","A","A","A"] ... 诸如此类
示例 3：
输入：tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
输出：16
解释：一种可能的解决方案是：
A -> B -> C -> A -> D -> E -> A -> F -> G -> A -> (待命) -> (待命) -> A -> (待命) -> (待命) -> A
提示：
1 <= task.length <= 104tasks[i]是大写英文字母n的取值范围为[0, 100]
**难度**: Medium
**标签**: 贪心算法、 队列、 数组、
# -*- coding: utf-8 -*-
# @Author : LG
"""
执行用时：44 ms, 在所有 Python3 提交中击败了100.00% 的用户
内存消耗：13.7 MB, 在所有 Python3 提交中击败了30.58% 的用户
解题思路：贪心算法
tasks = ["A","A","A","B","B","B","C","C","C","D","E"], n = 2, leastInterval = 16
A B C D
A B C E
A B C
tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2, leastInterval = 16
A B C
A D E
A F G
A _ _
A _ _
A
"""
class Solution:
def leastInterval(self, tasks: List[str], n: int) -> int:
count = collections.Counter(tasks) # 统计每个字符出现的次数
h = max([v for _,v in count.items()]) # 选最大的作为矩形高
w = len([v for _,v in count.items() if v == h]) # 将出现次数等于h的元素先放进矩阵，此时，矩阵宽w
ew = max(math.ceil((len(tasks) - h*w)/h), n-w+1) # 剩余字符放入矩阵，所需的额外宽ew
if n - w + 1 > (len(tasks) - h * w) / h: # 若剩余字符未放满ew*h
num = h * w + max((len(tasks) - h * w), ew * (h - 1))
else: # 若剩余字符超出了ew*h, h * w + len(tasks) - h * w
num = len(tasks)
return num
