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solution/3300-3399/3386.Button with Longest Push Time/README.md
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| comments: true | ||
| difficulty: 简单 | ||
| edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3386.Button%20with%20Longest%20Push%20Time/README.md | ||
| --- | ||
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| <!-- problem:start --> | ||
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| # [3386. 按下时间最长的按钮](https://leetcode.cn/problems/button-with-longest-push-time) | ||
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| [English Version](/solution/3300-3399/3386.Button%20with%20Longest%20Push%20Time/README_EN.md) | ||
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| ## 题目描述 | ||
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| <!-- description:start --> | ||
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| <p>给你一个二维数组 <code>events</code>，表示孩子在键盘上按下一系列按钮触发的按钮事件。</p> | ||
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| <p>每个 <code>events[i] = [index<sub>i</sub>, time<sub>i</sub>]</code> 表示在时间 <code>time<sub>i</sub></code> 时，按下了下标为 <code>index<sub>i</sub></code> 的按钮。</p> | ||
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| <ul> | ||
| <li>数组按照 <code>time</code> 的递增顺序<strong>排序</strong>。</li> | ||
| <li>按下一个按钮所需的时间是连续两次按钮按下的时间差。按下第一个按钮所需的时间就是其时间戳。</li> | ||
| </ul> | ||
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| <p>返回按下时间 <strong>最长 </strong>的按钮的 <code>index</code>。如果有多个按钮的按下时间相同，则返回 <code>index</code> 最小的按钮。</p> | ||
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| <p> </p> | ||
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| <p><strong class="example">示例 1：</strong></p> | ||
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| <div class="example-block"> | ||
| <p><strong>输入：</strong> <span class="example-io">events = [[1,2],[2,5],[3,9],[1,15]]</span></p> | ||
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| <p><strong>输出：</strong> <span class="example-io">1</span></p> | ||
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| <p><strong>解释：</strong></p> | ||
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| <ul> | ||
| <li>下标为 1 的按钮在时间 2 被按下。</li> | ||
| <li>下标为 2 的按钮在时间 5 被按下，因此按下时间为 <code>5 - 2 = 3</code>。</li> | ||
| <li>下标为 3 的按钮在时间 9 被按下，因此按下时间为 <code>9 - 5 = 4</code>。</li> | ||
| <li>下标为 1 的按钮再次在时间 15 被按下，因此按下时间为 <code>15 - 9 = 6</code>。</li> | ||
| </ul> | ||
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| <p>最终，下标为 1 的按钮按下时间最长，为 6。</p> | ||
| </div> | ||
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| <p><strong class="example">示例 2：</strong></p> | ||
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| <div class="example-block"> | ||
| <p><strong>输入：</strong> <span class="example-io">events = [[10,5],[1,7]]</span></p> | ||
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| <p><strong>输出：</strong> <span class="example-io">10</span></p> | ||
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| <p><strong>解释：</strong></p> | ||
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| <ul> | ||
| <li>下标为 10 的按钮在时间 5 被按下。</li> | ||
| <li>下标为 1 的按钮在时间 7 被按下，因此按下时间为 <code>7 - 5 = 2</code>。</li> | ||
| </ul> | ||
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| <p>最终，下标为 10 的按钮按下时间最长，为 5。</p> | ||
| </div> | ||
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| <p> </p> | ||
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| <p><strong>提示：</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= events.length <= 1000</code></li> | ||
| <li><code>events[i] == [index<sub>i</sub>, time<sub>i</sub>]</code></li> | ||
| <li><code>1 <= index<sub>i</sub>, time<sub>i</sub> <= 10<sup>5</sup></code></li> | ||
| <li>输入保证数组 <code>events</code> 按照 <code>time<sub>i</sub></code> 的递增顺序排序。</li> | ||
| </ul> | ||
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| <!-- description:end --> | ||
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| ## 解法 | ||
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| <!-- solution:start --> | ||
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| ### 方法一：一次遍历 | ||
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| 我们定义两个变量 $\textit{ans}$ 和 $t$，分别表示按下时间最长的按钮的索引和按下时间。 | ||
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| 接下来，我们从下标 $k = 1$ 开始遍历数组 $\textit{events}$，对于每个 $k$，我们计算当前按钮的按下时间 $d = t2 - t1$，其中 $t2$ 是当前按钮的按下时间，而 $t1$ 是前一个按钮的按下时间。如果 $d > t$ 或者 $d = t$ 且当前按钮的索引 $i$ 小于 $\textit{ans}$，我们更新 $\textit{ans} = i$ 和 $t = d$。 | ||
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| 最后，我们返回 $\textit{ans}$。 | ||
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| 时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{events}$ 的长度。空间复杂度 $O(1)$。 | ||
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| <!-- tabs:start --> | ||
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| #### Python3 | ||
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| ```python | ||
| class Solution: | ||
| def buttonWithLongestTime(self, events: List[List[int]]) -> int: | ||
| ans, t = events[0] | ||
| for (_, t1), (i, t2) in pairwise(events): | ||
| d = t2 - t1 | ||
| if d > t or (d == t and i < ans): | ||
| ans, t = i, d | ||
| return ans | ||
| ``` | ||
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| #### Java | ||
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| ```java | ||
| class Solution { | ||
| public int buttonWithLongestTime(int[][] events) { | ||
| int ans = events[0][0], t = events[0][1]; | ||
| for (int k = 1; k < events.length; ++k) { | ||
| int i = events[k][0], t2 = events[k][1], t1 = events[k - 1][1]; | ||
| int d = t2 - t1; | ||
| if (d > t || (d == t && ans > i)) { | ||
| ans = i; | ||
| t = d; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
| } | ||
| ``` | ||
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| #### C++ | ||
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| int buttonWithLongestTime(vector<vector<int>>& events) { | ||
| int ans = events[0][0], t = events[0][1]; | ||
| for (int k = 1; k < events.size(); ++k) { | ||
| int i = events[k][0], t2 = events[k][1], t1 = events[k - 1][1]; | ||
| int d = t2 - t1; | ||
| if (d > t || (d == t && ans > i)) { | ||
| ans = i; | ||
| t = d; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
| }; | ||
| ``` | ||
| #### Go | ||
| ```go | ||
| func buttonWithLongestTime(events [][]int) int { | ||
| ans, t := events[0][0], events[0][1] | ||
| for k, e := range events[1:] { | ||
| i, t2, t1 := e[0], e[1], events[k][1] | ||
| d := t2 - t1 | ||
| if d > t || (d == t && i < ans) { | ||
| ans, t = i, d | ||
| } | ||
| } | ||
| return ans | ||
| } | ||
| ``` | ||
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| #### TypeScript | ||
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| ```ts | ||
| function buttonWithLongestTime(events: number[][]): number { | ||
| let [ans, t] = events[0]; | ||
| for (let k = 1; k < events.length; ++k) { | ||
| const [i, t2] = events[k]; | ||
| const d = t2 - events[k - 1][1]; | ||
| if (d > t || (d === t && i < ans)) { | ||
| ans = i; | ||
| t = d; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
| ``` | ||
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| <!-- tabs:end --> | ||
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| <!-- solution:end --> | ||
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| <!-- problem:end --> |
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