There are n employees, each with a unique id from 0 to n - 1.
You are given a 2D integer array logs where logs[i] = [idi, leaveTimei] where:
idi is the id of the employee that worked on the ith task, andleaveTimei is the time at which the employee finished the ith task. All the values leaveTimei are unique.Note that the ith task starts the moment right after the (i - 1)th task ends, and the 0th task starts at time 0.
Return the id of the employee that worked the task with the longest time. If there is a tie between two or more employees, return the smallest id among them.
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Example 1:
\n\n\nInput: n = 10, logs = [[0,3],[2,5],[0,9],[1,15]]\nOutput: 1\nExplanation: \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\n\n
Example 2:
\n\n\nInput: n = 26, logs = [[1,1],[3,7],[2,12],[7,17]]\nOutput: 3\nExplanation: \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\n\n
Example 3:
\n\n\nInput: n = 2, logs = [[0,10],[1,20]]\nOutput: 0\nExplanation: \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\n\n
\n
Constraints:
\n\n2 <= n <= 5001 <= logs.length <= 500logs[i].length == 20 <= idi <= n - 11 <= leaveTimei <= 500idi != idi+1leaveTimei are sorted in a strictly increasing order.共有 n 位员工，每位员工都有一个从 0 到 n - 1 的唯一 id 。
给你一个二维整数数组 logs ，其中 logs[i] = [idi, leaveTimei] ：
idi 是处理第 i 个任务的员工的 id ，且leaveTimei 是员工完成第 i 个任务的时刻。所有 leaveTimei 的值都是 唯一 的。注意，第 i 个任务在第 (i - 1) 个任务结束后立即开始，且第 0 个任务从时刻 0 开始。
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 最小 的 id 。
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示例 1：
\n\n\n输入：n = 10, logs = [[0,3],[2,5],[0,9],[1,15]]\n输出：1\n解释：\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\n\n
示例 2：
\n\n\n输入：n = 26, logs = [[1,1],[3,7],[2,12],[7,17]]\n输出：3\n解释：\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\n\n
示例 3：
\n\n\n输入：n = 2, logs = [[0,10],[1,20]]\n输出：0\n解释：\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\n\n
\n\n
提示：
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