{ "data": { "question": { "questionId": "621", "questionFrontendId": "621", "categoryTitle": "Algorithms", "boundTopicId": 1392, "title": "Task Scheduler", "titleSlug": "task-scheduler", "content": "

You are given an array of CPU tasks, each labeled with a letter from A to Z, and a number n. Each CPU interval can be idle or allow the completion of one task. Tasks can be completed in any order, but there's a constraint: there has to be a gap of at least n intervals between two tasks with the same label.

\n\n

Return the minimum number of CPU intervals required to complete all tasks.

\n\n

 

\n

Example 1:

\n\n
\n

Input: tasks = ["A","A","A","B","B","B"], n = 2

\n\n

Output: 8

\n\n

Explanation: A possible sequence is: A -> B -> idle -> A -> B -> idle -> A -> B.

\n\n

After completing task A, you must wait two intervals before doing A again. The same applies to task B. In the 3rd interval, neither A nor B can be done, so you idle. By the 4th interval, you can do A again as 2 intervals have passed.

\n
\n\n

Example 2:

\n\n
\n

Input: tasks = ["A","C","A","B","D","B"], n = 1

\n\n

Output: 6

\n\n

Explanation: A possible sequence is: A -> B -> C -> D -> A -> B.

\n\n

With a cooling interval of 1, you can repeat a task after just one other task.

\n
\n\n

Example 3:

\n\n
\n

Input: tasks = ["A","A","A", "B","B","B"], n = 3

\n\n

Output: 10

\n\n

Explanation: A possible sequence is: A -> B -> idle -> idle -> A -> B -> idle -> idle -> A -> B.

\n\n

There are only two types of tasks, A and B, which need to be separated by 3 intervals. This leads to idling twice between repetitions of these tasks.

\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "任务调度器", "translatedContent": "

给你一个用字符数组 tasks 表示的 CPU 需要执行的任务列表，用字母 A 到 Z 表示，以及一个冷却时间 n。每个周期或时间间隔允许完成一项任务。任务可以按任何顺序完成，但有一个限制：两个 相同种类 的任务之间必须有长度为 n 的冷却时间。

\n\n

返回完成所有任务所需要的 最短时间间隔 。

\n\n

 

\n\n

示例 1：

\n\n
输入：tasks = [\"A\",\"A\",\"A\",\"B\",\"B\",\"B\"], n = 2
\n\n
输出：8
\n\n
解释：
\n\n
在完成任务 A 之后，你必须等待两个间隔。对任务 B 来说也是一样。在第 3 个间隔，A 和 B 都不能完成，所以你需要待命。在第 4 个间隔，由于已经经过了 2 个间隔，你可以再次执行 A 任务。
\n\n
 
\n\n

示例 2：

\n\n
\n

输入：tasks = [\"A\",\"C\",\"A\",\"B\",\"D\",\"B\"], n = 1

\n\n

输出：6

\n\n

解释：一种可能的序列是：A -> B -> C -> D -> A -> B。

\n\n

由于冷却间隔为 1，你可以在完成另一个任务后重复执行这个任务。

\n
\n\n

示例 3：

\n\n
输入：tasks = [\"A\",\"A\",\"A\",\"B\",\"B\",\"B\"], n = 3
\n\n
输出：10
\n\n
解释：一种可能的序列为：A -> B -> idle -> idle -> A -> B -> idle -> idle -> A -> B。
\n\n
只有两种任务类型，A 和 B，需要被 3 个间隔分割。这导致重复执行这些任务的间隔当中有两次待命状态。
\n\n

 

\n\n

提示：

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\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"
Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n
\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"
\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n
\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a-O2 --disable-reflection<\\/code><\\/p>\\r\\n\\r\\n
\\u5feb\\u901f\\u5165\\u95e8\\u8bf7\\u67e5\\u9605\\u300c\\u4ed3\\u9889\\u7f16\\u7a0b\\u8bed\\u8a00\\u5f00\\u53d1\\u6307\\u5357\\u300d<\\/a><\\/p>\"]}",
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            "style": "LEETCODE",
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}