leetcode-problemset/leetcode-cn/originData/single-threaded-cpu.json

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            "questionId": "1962",
            "questionFrontendId": "1834",
            "categoryTitle": "Algorithms",
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            "title": "Single-Threaded CPU",
            "titleSlug": "single-threaded-cpu",
            "content": "<p>You are given <code>n</code> tasks labeled from <code>0</code> to <code>n - 1</code> represented by a 2D integer array <code>tasks</code>, where <code>tasks[i] = [enqueueTime<sub>i</sub>, processingTime<sub>i</sub>]</code> means that the <code>i<sup>th</sup></code> task will be available to process at <code>enqueueTime<sub>i</sub></code> and will take <code>processingTime<sub>i</sub></code><sub> </sub>to finish processing.</p>\n\n<p>You have a single-threaded CPU that can process <strong>at most one</strong> task at a time and will act in the following way:</p>\n\n<ul>\n\t<li>If the CPU is idle and there are no available tasks to process, the CPU remains idle.</li>\n\t<li>If the CPU is idle and there are available tasks, the CPU will choose the one with the <strong>shortest processing time</strong>. If multiple tasks have the same shortest processing time, it will choose the task with the smallest index.</li>\n\t<li>Once a task is started, the CPU will <strong>process the entire task</strong> without stopping.</li>\n\t<li>The CPU can finish a task then start a new one instantly.</li>\n</ul>\n\n<p>Return <em>the order in which the CPU will process the tasks.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> tasks = [[1,2],[2,4],[3,2],[4,1]]\n<strong>Output:</strong> [0,2,3,1]\n<strong>Explanation: </strong>The events go as follows: \n- At time = 1, task 0 is available to process. Available tasks = {0}.\n- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.\n- At time = 2, task 1 is available to process. Available tasks = {1}.\n- At time = 3, task 2 is available to process. Available tasks = {1, 2}.\n- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.\n- At time = 4, task 3 is available to process. Available tasks = {1, 3}.\n- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.\n- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.\n- At time = 10, the CPU finishes task 1 and becomes idle.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]\n<strong>Output:</strong> [4,3,2,0,1]\n<strong>Explanation</strong><strong>: </strong>The events go as follows:\n- At time = 7, all the tasks become available. Available tasks = {0,1,2,3,4}.\n- Also at time = 7, the idle CPU starts processing task 4. Available tasks = {0,1,2,3}.\n- At time = 9, the CPU finishes task 4 and starts processing task 3. Available tasks = {0,1,2}.\n- At time = 13, the CPU finishes task 3 and starts processing task 2. Available tasks = {0,1}.\n- At time = 18, the CPU finishes task 2 and starts processing task 0. Available tasks = {1}.\n- At time = 28, the CPU finishes task 0 and starts processing task 1. Available tasks = {}.\n- At time = 40, the CPU finishes task 1 and becomes idle.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>tasks.length == n</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= enqueueTime<sub>i</sub>, processingTime<sub>i</sub> &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "单线程 CPU",
            "translatedContent": "<p>给你一个二维数组 <code>tasks</code> ，用于表示 <code>n</code> 项从 <code>0</code> 到 <code>n - 1</code> 编号的任务。其中 <code>tasks[i] = [enqueueTime<sub>i</sub>, processingTime<sub>i</sub>]</code> 意味着第 <code>i<sup></sup></code> 项任务将会于 <code>enqueueTime<sub>i</sub></code> 时进入任务队列，需要 <code>processingTime<sub>i</sub></code><sub> </sub>的时长完成执行。</p>\n\n<p>现有一个单线程 CPU ，同一时间只能执行 <strong>最多一项</strong> 任务，该 CPU 将会按照下述方式运行：</p>\n\n<ul>\n\t<li>如果 CPU 空闲，且任务队列中没有需要执行的任务，则 CPU 保持空闲状态。</li>\n\t<li>如果 CPU 空闲，但任务队列中有需要执行的任务，则 CPU 将会选择 <strong>执行时间最短</strong> 的任务开始执行。如果多个任务具有同样的最短执行时间，则选择下标最小的任务开始执行。</li>\n\t<li>一旦某项任务开始执行，CPU 在 <strong>执行完整个任务</strong> 前都不会停止。</li>\n\t<li>CPU 可以在完成一项任务后，立即开始执行一项新任务。</li>\n</ul>\n\n<p>返回<em> </em>CPU<em> </em>处理任务的顺序。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>tasks = [[1,2],[2,4],[3,2],[4,1]]\n<strong>输出：</strong>[0,2,3,1]\n<strong>解释：</strong>事件按下述流程运行： \n- time = 1 ，任务 0 进入任务队列，可执行任务项 = {0}\n- 同样在 time = 1 ，空闲状态的 CPU 开始执行任务 0 ，可执行任务项 = {}\n- time = 2 ，任务 1 进入任务队列，可执行任务项 = {1}\n- time = 3 ，任务 2 进入任务队列，可执行任务项 = {1, 2}\n- 同样在 time = 3 ，CPU 完成任务 0 并开始执行队列中用时最短的任务 2 ，可执行任务项 = {1}\n- time = 4 ，任务 3 进入任务队列，可执行任务项 = {1, 3}\n- time = 5 ，CPU 完成任务 2 并开始执行队列中用时最短的任务 3 ，可执行任务项 = {1}\n- time = 6 ，CPU 完成任务 3 并开始执行任务 1 ，可执行任务项 = {}\n- time = 10 ，CPU 完成任务 1 并进入空闲状态\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]\n<strong>输出：</strong>[4,3,2,0,1]\n<strong>解释：</strong>事件按下述流程运行： \n- time = 7 ，所有任务同时进入任务队列，可执行任务项  = {0,1,2,3,4}\n- 同样在 time = 7 ，空闲状态的 CPU 开始执行任务 4 ，可执行任务项 = {0,1,2,3}\n- time = 9 ，CPU 完成任务 4 并开始执行任务 3 ，可执行任务项 = {0,1,2}\n- time = 13 ，CPU 完成任务 3 并开始执行任务 2 ，可执行任务项 = {0,1}\n- time = 18 ，CPU 完成任务 2 并开始执行任务 0 ，可执行任务项 = {1}\n- time = 28 ，CPU 完成任务 0 并开始执行任务 1 ，可执行任务项 = {}\n- time = 40 ，CPU 完成任务 1 并进入空闲状态</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>tasks.length == n</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= enqueueTime<sub>i</sub>, processingTime<sub>i</sub> &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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\\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.11<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 3.3.1<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u4f7f\\u7528\\u7684\\u662f JetBrains \\u63d0\\u4f9b\\u7684 experimental compiler\\u3002\\u5982\\u679c\\u60a8\\u8ba4\\u4e3a\\u60a8\\u9047\\u5230\\u4e86\\u7f16\\u8bd1\\u5668\\u76f8\\u5173\\u7684\\u95ee\\u9898\\uff0c\\u8bf7\\u5411\\u6211\\u4eec\\u53cd\\u9988<\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.79.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a> \\u548c <a href=\\\"https:\\/\\/crates.io\\/crates\\/regex\\\" target=\\\"_blank\\\">regex<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\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\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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