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feat: add solutions to lc problems: No.4020~4022
Co-authored-by: Cursor <cursoragent@cursor.com>
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‎solution/4000-4099/4020.Elevator Requests I/README.md‎

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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4020.Elevator%20Requests%20I/README.md
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---
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<!-- problem:start -->
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# [4020. 电梯请求 I](https://leetcode.cn/problems/elevator-requests-i)
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[English Version](/solution/4000-4099/4020.Elevator%20Requests%20I/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你一个整数 <code>n</code> ，表示一栋楼房的楼层数，楼层编号从 0 到 <code>n - 1</code> 。</p>
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<p>同时给你一个整数数组 <code>requests</code> ，其中 <code>requests</code> 表示楼层请求的序列。</p>
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<p>一部电梯初始在 0 层，遵循以下规则：</p>
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<ul>
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<li>电梯每秒移动一层。</li>
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<li>电梯按给定的顺序处理请求。</li>
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<li>如果电梯已经在请求的楼层，则不需要移动。</li>
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<li>处理完一个请求后，电梯立即开始向下一个请求的楼层移动。</li>
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</ul>
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<p>返回处理所有请求所需的 <strong>总时间</strong> （以秒为单位）。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">n = 5, requests = [2,1,4,3]</span></p>
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<p><strong>输出：</strong> <span class="example-io">7</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li><code>requests[0] = 2</code>：从 0 层移动到 2 层需要 2 秒。</li>
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<li><code>requests[1] = 1</code>：从 2 层移动到 1 层需要 1 秒。</li>
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<li><code>requests[2] = 4</code>：从 1 层移动到 4 层需要 3 秒。</li>
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<li><code>requests[3] = 3</code>：从 4 层移动到 3 层需要 1 秒。</li>
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</ul>
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<p>所需的总时间是 <code>2 + 1 + 3 + 1 = 7</code> 秒。</p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">n = 3, requests = [2,0,0]</span></p>
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<p><strong>输出：</strong> <span class="example-io">4</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li><code>requests[0] = 2</code>：从 0 层移动到 2 层需要 2 秒。</li>
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<li><code>requests[1] = 0</code>：从 2 层移动到 0 层需要 2 秒。</li>
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<li><code>requests[2] = 0</code>：不需要移动。</li>
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</ul>
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<p>所需的总时间是 <code>2 + 2 + 0 = 4</code> 秒。</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>1 &lt;= requests.length &lt;= 100</code></li>
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<li><code>0 &lt;= requests[i] &lt;= n - 1</code></li>
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</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：模拟
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电梯从 $0$ 层出发，按给定顺序处理请求。相邻两次请求之间的移动时间为两层楼层编号之差的绝对值。第一个请求从 $0$ 层到 $\textit{requests}[0]$，耗时即为 $\textit{requests}[0]$；之后将相邻请求的楼层差绝对值累加即可。
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时间复杂度 $O(m)$，空间复杂度 $O(1)$。其中 $m$ 是请求的数量。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def elevatorRequests(self, n: int, requests: list[int]) -> int:
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return requests[0] + sum(abs(x - y) for x, y in pairwise(requests))
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```
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#### Java
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```java
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class Solution {
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public int elevatorRequests(int n, int[] requests) {
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int ans = requests[0];
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for (int i = 1; i < requests.length; ++i) {
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ans += Math.abs(requests[i - 1] - requests[i]);
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}
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return ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int elevatorRequests(int n, vector<int>& requests) {
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int ans = requests[0];
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for (int i = 1; i < requests.size(); ++i) {
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ans += abs(requests[i - 1] - requests[i]);
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func elevatorRequests(n int, requests []int) int {
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ans := requests[0]
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for i, x := range requests[1:] {
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ans += abs(x - requests[i])
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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#### TypeScript
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```ts
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function elevatorRequests(n: number, requests: number[]): number {
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let ans: number = requests[0];
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for (let i = 1; i < requests.length; ++i) {
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ans += Math.abs(requests[i] - requests[i - 1]);
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/4000-4099/4020.Elevator Requests I/README_EN.md‎

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---
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comments: true
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difficulty: Easy
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4020.Elevator%20Requests%20I/README_EN.md
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---
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<!-- problem:start -->
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# [4020. Elevator Requests I](https://leetcode.com/problems/elevator-requests-i)
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[中文文档](/solution/4000-4099/4020.Elevator%20Requests%20I/README.md)
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## Description
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<!-- description:start -->
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<p>You are given an integer <code>n</code> denoting the number of floors in a building, where the floors are numbered from 0 to <code>n - 1</code>.</p>
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<p>You are also given an integer array <code>requests</code>, where <code>requests</code> represents the sequence of floor requests.</p>
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<p>An elevator starts at floor 0, and follows these rules:</p>
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<ul>
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<li>The elevator moves one floor per second.</li>
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<li>The elevator serves requests in the given order.</li>
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<li>If the elevator is already on the requested floor, no movement is needed.</li>
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<li>After serving a request, the elevator immediately starts moving toward the next request.</li>
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</ul>
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<p>Return the <strong>total time</strong> (in seconds) required to serve all requests.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 5, requests = [2,1,4,3]</span></p>
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<p><strong>Output:</strong> <span class="example-io">7</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li><code>requests[0] = 2</code>: Moving from floor 0 to floor 2 takes 2 seconds.</li>
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<li><code>requests[1] = 1</code>: Moving from floor 2 to floor 1 takes 1 second.</li>
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<li><code>requests[2] = 4</code>: Moving from floor 1 to floor 4 takes 3 seconds.</li>
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<li><code>requests[3] = 3</code>: Moving from floor 4 to floor 3 takes 1 second.</li>
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</ul>
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<p>The total time required is <code>2 + 1 + 3 + 1 = 7</code> seconds.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 3, requests = [2,0,0]</span></p>
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<p><strong>Output:</strong> <span class="example-io">4</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li><code>requests[0] = 2</code>: Moving from floor 0 to floor 2 takes 2 seconds.</li>
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<li><code>requests[1] = 0</code>: Moving from floor 2 to floor 0 takes 2 seconds.</li>
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<li><code>requests[2] = 0</code>: No movement is needed.</li>
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</ul>
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<p>The total time required is <code>2 + 2 + 0 = 4</code> seconds.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>1 &lt;= requests.length &lt;= 100</code></li>
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<li><code>0 &lt;= requests[i] &lt;= n - 1</code></li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Simulation
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The elevator starts at floor $0$ and serves requests in the given order. The travel time between two consecutive requests is the absolute difference of their floor numbers. The first request goes from floor $0$ to $\textit{requests}[0]$, which takes $\textit{requests}[0]$ seconds. Then we add the absolute differences of adjacent requests.
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The time complexity is $O(m)$, and the space complexity is $O(1)$, where $m$ is the number of requests.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def elevatorRequests(self, n: int, requests: list[int]) -> int:
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return requests[0] + sum(abs(x - y) for x, y in pairwise(requests))
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```
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#### Java
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```java
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class Solution {
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public int elevatorRequests(int n, int[] requests) {
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int ans = requests[0];
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for (int i = 1; i < requests.length; ++i) {
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ans += Math.abs(requests[i - 1] - requests[i]);
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}
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return ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int elevatorRequests(int n, vector<int>& requests) {
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int ans = requests[0];
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for (int i = 1; i < requests.size(); ++i) {
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ans += abs(requests[i - 1] - requests[i]);
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func elevatorRequests(n int, requests []int) int {
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ans := requests[0]
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for i, x := range requests[1:] {
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ans += abs(x - requests[i])
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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#### TypeScript
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```ts
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function elevatorRequests(n: number, requests: number[]): number {
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let ans: number = requests[0];
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for (let i = 1; i < requests.length; ++i) {
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ans += Math.abs(requests[i] - requests[i - 1]);
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/4000-4099/4020.Elevator Requests I/Solution.cpp‎

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class Solution {
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public:
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int elevatorRequests(int n, vector<int>& requests) {
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int ans = requests[0];
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for (int i = 1; i < requests.size(); ++i) {
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ans += abs(requests[i - 1] - requests[i]);
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}
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return ans;
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}
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};

‎solution/4000-4099/4020.Elevator Requests I/Solution.go‎

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func elevatorRequests(n int, requests []int) int {
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ans := requests[0]
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for i, x := range requests[1:] {
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ans += abs(x - requests[i])
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}

‎solution/4000-4099/4020.Elevator Requests I/Solution.java‎

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class Solution {
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public int elevatorRequests(int n, int[] requests) {
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int ans = requests[0];
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for (int i = 1; i < requests.length; ++i) {
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ans += Math.abs(requests[i - 1] - requests[i]);
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}
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return ans;
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}
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}

‎solution/4000-4099/4020.Elevator Requests I/Solution.py‎

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class Solution:
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def elevatorRequests(self, n: int, requests: list[int]) -> int:
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return requests[0] + sum(abs(x - y) for x, y in pairwise(requests))

‎solution/4000-4099/4020.Elevator Requests I/Solution.ts‎

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function elevatorRequests(n: number, requests: number[]): number {
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let ans: number = requests[0];
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for (let i = 1; i < requests.length; ++i) {
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ans += Math.abs(requests[i] - requests[i - 1]);
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}
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return ans;
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}

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