You are given a 2D array events which represents a sequence of events where a child pushes a series of buttons on a keyboard.
Each events[i] = [indexi, timei] indicates that the button at index indexi was pressed at time timei.
time.Return the index of the button that took the longest time to push. If multiple buttons have the same longest time, return the button with the smallest index.
\n
Example 1:
\n\nInput: events = [[1,2],[2,5],[3,9],[1,15]]
\n\nOutput: 1
\n\nExplanation:
\n\n5 - 2 = 3 units of time.9 - 5 = 4 units of time.15 - 9 = 6 units of time.Example 2:
\n\nInput: events = [[10,5],[1,7]]
\n\nOutput: 10
\n\nExplanation:
\n\n7 - 5 = 2 units of time.\n
Constraints:
\n\n1 <= events.length <= 1000events[i] == [indexi, timei]1 <= indexi, timei <= 105events is sorted in increasing order of timei.给你一个二维数组 events，表示孩子在键盘上按下一系列按钮触发的按钮事件。
每个 events[i] = [indexi, timei] 表示在时间 timei 时，按下了下标为 indexi 的按钮。
time 的递增顺序排序。返回按下时间 最长 的按钮的 index。如果有多个按钮的按下时间相同，则返回 index 最小的按钮。
\n\n
示例 1：
\n\n输入： events = [[1,2],[2,5],[3,9],[1,15]]
\n\n输出： 1
\n\n解释：
\n\n5 - 2 = 3。9 - 5 = 4。15 - 9 = 6。最终，下标为 1 的按钮按下时间最长，为 6。
\n示例 2：
\n\n输入： events = [[10,5],[1,7]]
\n\n输出： 10
\n\n解释：
\n\n7 - 5 = 2。最终，下标为 10 的按钮按下时间最长，为 5。
\n\n\n
提示：
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