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Commit 41233e3

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feat: add weekly contest 446 (#4362)
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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3522.Calculate%20Score%20After%20Performing%20Instructions/README.md
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---
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<!-- problem:start -->
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# [3522. 执行指令后的得分](https://leetcode.cn/problems/calculate-score-after-performing-instructions)
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[English Version](/solution/3500-3599/3522.Calculate%20Score%20After%20Performing%20Instructions/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你两个数组:<code>instructions</code> 和 <code>values</code>,数组的长度均为 <code>n</code>。</p>
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<p>你需要根据以下规则模拟一个过程:</p>
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<ul>
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<li>从下标&nbsp;<code>i = 0</code> 的第一个指令开始,初始得分为 0。</li>
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<li>如果 <code>instructions[i]</code> 是 <code>"add"</code>:
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<ul>
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<li>将 <code>values[i]</code> 加到你的得分中。</li>
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<li>移动到下一个指令 <code>(i + 1)</code>。</li>
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</ul>
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</li>
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<li>如果 <code>instructions[i]</code> 是 <code>"jump"</code>:
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<ul>
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<li>移动到下标为&nbsp;<code>(i + values[i])</code> 的指令,但不修改你的得分。</li>
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</ul>
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</li>
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</ul>
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<p>当以下任一情况发生时,过程会终止:</p>
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<ul>
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<li>越界(即 <code>i &lt; 0</code> 或 <code>i &gt;= n</code>),或</li>
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<li>尝试再次执行已经执行过的指令。被重复访问的指令不会再次执行。</li>
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</ul>
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<p>返回过程结束时的得分。</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">instructions = ["jump","add","add","jump","add","jump"], values = [2,1,3,1,-2,-3]</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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<p>从下标&nbsp;0 开始模拟过程:</p>
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<ul>
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<li>下标 0:指令是 <code>"jump"</code>,移动到下标&nbsp;<code>0 + 2 = 2</code>。</li>
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<li>下标 2:指令是 <code>"add"</code>,将 <code>values[2] = 3</code> 加到得分中,移动到下标&nbsp;3。得分变为 3。</li>
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<li>下标 3:指令是 <code>"jump"</code>,移动到下标&nbsp;<code>3 + 1 = 4</code>。</li>
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<li>下标 4:指令是 <code>"add"</code>,将 <code>values[4] = -2</code> 加到得分中,移动到下标&nbsp;5。得分变为 1。</li>
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<li>下标 5:指令是 <code>"jump"</code>,移动到下标&nbsp;<code>5 + (-3) = 2</code>。</li>
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<li>下标 2:已经访问过。过程结束。</li>
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</ul>
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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">instructions = ["jump","add","add"], values = [3,1,1]</span></p>
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<p><strong>输出:</strong> <span class="example-io">0</span></p>
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<p><strong>解释:</strong></p>
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<p>从下标&nbsp;0 开始模拟过程:</p>
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<ul>
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<li>下标 0:指令是 <code>"jump"</code>,移动到下标&nbsp;<code>0 + 3 = 3</code>。</li>
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<li>下标 3:越界。过程结束。</li>
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</ul>
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</div>
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<p><strong class="example">示例 3:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong> <span class="example-io">instructions = ["jump"], values = [0]</span></p>
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<p><strong>输出:</strong> <span class="example-io">0</span></p>
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<p><strong>解释:</strong></p>
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<p>从下标&nbsp;0 开始模拟过程:</p>
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<ul>
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<li>下标 0:指令是 <code>"jump"</code>,移动到下标&nbsp;<code>0 + 0 = 0</code>。</li>
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<li>下标 0:已经访问过。过程结束。</li>
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</ul>
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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>n == instructions.length == values.length</code></li>
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<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>instructions[i]</code> 只能是 <code>"add"</code> 或 <code>"jump"</code>。</li>
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<li><code>-10<sup>5</sup> &lt;= values[i] &lt;= 10<sup>5</sup></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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我们根据题意模拟即可。
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我们定义一个长度为 $n$ 的布尔数组 $\textit{vis},ドル用于记录每一条指令是否被执行过,初始时均为 $\text{false}$。
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然后我们从下标 $i = 0$ 开始,循环执行以下操作:
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1. 将 $\textit{vis}[i]$ 置为 $\text{true}$;
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2. 如果 $\textit{instructions}[i]$ 的第一个字符为 'a',那么我们将答案增加 $\textit{value}[i],ドル然后 $i$ 加 1ドル$;否则,我们将 $i$ 增加 $\textit{value}[i]$。
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循环,直至 $i \lt 0$ 或者 $i \ge n,ドル或者 $\textit{vis}[i]$ 为 $\text{true}$。
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最后返回答案即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{value}$ 的长度。
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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 calculateScore(self, instructions: List[str], values: List[int]) -> int:
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n = len(values)
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vis = [False] * n
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ans = i = 0
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while 0 <= i < n and not vis[i]:
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vis[i] = True
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if instructions[i][0] == "a":
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ans += values[i]
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i += 1
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else:
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i = i + values[i]
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return ans
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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 long calculateScore(String[] instructions, int[] values) {
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int n = values.length;
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boolean[] vis = new boolean[n];
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long ans = 0;
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int i = 0;
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while (i >= 0 && i < n && !vis[i]) {
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vis[i] = true;
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if (instructions[i].charAt(0) == 'a') {
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ans += values[i];
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i += 1;
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} else {
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i = i + values[i];
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}
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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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long long calculateScore(vector<string>& instructions, vector<int>& values) {
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int n = values.size();
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vector<bool> vis(n, false);
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long long ans = 0;
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int i = 0;
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while (i >= 0 && i < n && !vis[i]) {
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vis[i] = true;
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if (instructions[i][0] == 'a') {
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ans += values[i];
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i += 1;
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} else {
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i += values[i];
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}
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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 calculateScore(instructions []string, values []int) (ans int64) {
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n := len(values)
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vis := make([]bool, n)
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i := 0
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for i >= 0 && i < n && !vis[i] {
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vis[i] = true
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if instructions[i][0] == 'a' {
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ans += int64(values[i])
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i += 1
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} else {
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i += values[i]
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}
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}
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return
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}
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```
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#### TypeScript
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```ts
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function calculateScore(instructions: string[], values: number[]): number {
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const n = values.length;
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const vis: boolean[] = Array(n).fill(false);
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let ans = 0;
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let i = 0;
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while (i >= 0 && i < n && !vis[i]) {
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vis[i] = true;
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if (instructions[i][0] === 'a') {
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ans += values[i];
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i += 1;
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} else {
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i += values[i];
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}
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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 -->

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