leetcode-problemset/leetcode-cn/originData/the-employee-that-worked-on-the-longest-task.json

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            "questionId": "2518",
            "questionFrontendId": "2432",
            "categoryTitle": "Algorithms",
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            "title": "The Employee That Worked on the Longest Task",
            "titleSlug": "the-employee-that-worked-on-the-longest-task",
            "content": "<p>There are <code>n</code> employees, each with a unique id from <code>0</code> to <code>n - 1</code>.</p>\n\n<p>You are given a 2D integer array <code>logs</code> where <code>logs[i] = [id<sub>i</sub>, leaveTime<sub>i</sub>]</code> where:</p>\n\n<ul>\n\t<li><code>id<sub>i</sub></code> is the id of the employee that worked on the <code>i<sup>th</sup></code> task, and</li>\n\t<li><code>leaveTime<sub>i</sub></code> is the time at which the employee finished the <code>i<sup>th</sup></code> task. All the values <code>leaveTime<sub>i</sub></code> are <strong>unique</strong>.</li>\n</ul>\n\n<p>Note that the <code>i<sup>th</sup></code> task starts the moment right after the <code>(i - 1)<sup>th</sup></code> task ends, and the <code>0<sup>th</sup></code> task starts at time <code>0</code>.</p>\n\n<p>Return <em>the id of the employee that worked the task with the longest time.</em> If there is a tie between two or more employees, return<em> the <strong>smallest</strong> id among them</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 10, logs = [[0,3],[2,5],[0,9],[1,15]]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> \nTask 0 started at 0 and ended at 3 with 3 units of times.\nTask 1 started at 3 and ended at 5 with 2 units of times.\nTask 2 started at 5 and ended at 9 with 4 units of times.\nTask 3 started at 9 and ended at 15 with 6 units of times.\nThe task with the longest time is task 3 and the employee with id 1 is the one that worked on it, so we return 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 26, logs = [[1,1],[3,7],[2,12],[7,17]]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> \nTask 0 started at 0 and ended at 1 with 1 unit of times.\nTask 1 started at 1 and ended at 7 with 6 units of times.\nTask 2 started at 7 and ended at 12 with 5 units of times.\nTask 3 started at 12 and ended at 17 with 5 units of times.\nThe tasks with the longest time is task 1. The employee that worked on it is 3, so we return 3.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2, logs = [[0,10],[1,20]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> \nTask 0 started at 0 and ended at 10 with 10 units of times.\nTask 1 started at 10 and ended at 20 with 10 units of times.\nThe tasks with the longest time are tasks 0 and 1. The employees that worked on them are 0 and 1, so we return the smallest id 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 500</code></li>\n\t<li><code>1 &lt;= logs.length &lt;= 500</code></li>\n\t<li><code>logs[i].length == 2</code></li>\n\t<li><code>0 &lt;= id<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>1 &lt;= leaveTime<sub>i</sub> &lt;= 500</code></li>\n\t<li><code>id<sub>i</sub> != id<sub>i+1</sub></code></li>\n\t<li><code>leaveTime<sub>i</sub></code> are sorted in a strictly increasing order.</li>\n</ul>\n",
            "translatedTitle": "处理用时最长的那个任务的员工",
            "translatedContent": "<p>共有 <code>n</code> 位员工，每位员工都有一个从 <code>0</code> 到 <code>n - 1</code> 的唯一 id 。</p>\n\n<p>给你一个二维整数数组 <code>logs</code> ，其中 <code>logs[i] = [id<sub>i</sub>, leaveTime<sub>i</sub>]</code> ：</p>\n\n<ul>\n\t<li><code>id<sub>i</sub></code> 是处理第 <code>i</code> 个任务的员工的 id ，且</li>\n\t<li><code>leaveTime<sub>i</sub></code> 是员工完成第 <code>i</code> 个任务的时刻。所有 <code>leaveTime<sub>i</sub></code> 的值都是 <strong>唯一</strong> 的。</li>\n</ul>\n\n<p>注意，第 <code>i</code> 个任务在第 <code>(i - 1)</code> 个任务结束后立即开始，且第 <code>0</code> 个任务从时刻 <code>0</code> 开始。</p>\n\n<p>返回处理用时最长的那个任务的员工的 id 。如果存在两个或多个员工同时满足，则返回几人中 <strong>最小</strong> 的 id 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 10, logs = [[0,3],[2,5],[0,9],[1,15]]\n<strong>输出：</strong>1\n<strong>解释：</strong>\n任务 0 于时刻 0 开始，且在时刻 3 结束，共计 3 个单位时间。\n任务 1 于时刻 3 开始，且在时刻 5 结束，共计 2 个单位时间。\n任务 2 于时刻 5 开始，且在时刻 9 结束，共计 4 个单位时间。\n任务 3 于时刻 9 开始，且在时刻 15 结束，共计 6 个单位时间。\n时间最长的任务是任务 3 ，而 id 为 1 的员工是处理此任务的员工，所以返回 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 26, logs = [[1,1],[3,7],[2,12],[7,17]]\n<strong>输出：</strong>3\n<strong>解释：</strong>\n任务 0 于时刻 0 开始，且在时刻 1 结束，共计 1 个单位时间。\n任务 1 于时刻 1 开始，且在时刻 7 结束，共计 6 个单位时间。\n任务 2 于时刻 7 开始，且在时刻 12 结束，共计 5 个单位时间。\n任务 3 于时刻 12 开始，且在时刻 17 结束，共计 5 个单位时间。\n时间最长的任务是任务 1 ，而 id 为 3 的员工是处理此任务的员工，所以返回 3 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 2, logs = [[0,10],[1,20]]\n<strong>输出：</strong>0\n<strong>解释：</strong>\n任务 0 于时刻 0 开始，且在时刻 10 结束，共计 10 个单位时间。\n任务 1 于时刻 10 开始，且在时刻 20 结束，共计 10 个单位时间。\n时间最长的任务是任务 0 和 1 ，处理这两个任务的员工的 id 分别是 0 和 1 ，所以返回最小的 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 500</code></li>\n\t<li><code>1 &lt;= logs.length &lt;= 500</code></li>\n\t<li><code>logs[i].length == 2</code></li>\n\t<li><code>0 &lt;= id<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>1 &lt;= leaveTime<sub>i</sub> &lt;= 500</code></li>\n\t<li><code>id<sub>i</sub> != id<sub>i + 1</sub></code></li>\n\t<li><code>leaveTime<sub>i</sub></code> 按严格递增顺序排列</li>\n</ul>\n",
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                "Find the time of the longest task",
                "Store each employee’s longest task time in a hash table",
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