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feat: add biweekly contest 131 #2915

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feat: add biweekly contest 131
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107 changes: 107 additions & 0 deletions solution/3100-3199/3158.Find the XOR of Numbers Which Appear Twice/README.md
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---
comments: true
difficulty: 简单
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3158.Find%20the%20XOR%20of%20Numbers%20Which%20Appear%20Twice/README.md
---

<!-- problem:start -->

# [3158. 求出出现两次数字的 XOR 值](https://leetcode.cn/problems/find-the-xor-of-numbers-which-appear-twice)

[English Version](/solution/3100-3199/3158.Find%20the%20XOR%20of%20Numbers%20Which%20Appear%20Twice/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给你一个数组&nbsp;<code>nums</code>&nbsp;，数组中的数字 <strong>要么</strong> 出现一次，<strong>要么</strong>&nbsp;出现两次。</p>

<p>请你返回数组中所有出现两次数字的按位<em>&nbsp;</em><code>XOR</code>&nbsp;值，如果没有数字出现过两次，返回 0 。</p>

<p>&nbsp;</p>

<p><strong class="example">示例 1：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>nums = [1,2,1,3]</span></p>

<p><span class="example-io"><b>输出：</b>1</span></p>

<p><strong>解释：</strong></p>

<p><code>nums</code>&nbsp;中唯一出现过两次的数字是 1 。</p>
</div>

<p><strong class="example">示例 2：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>nums = [1,2,3]</span></p>

<p><span class="example-io"><b>输出：</b>0</span></p>

<p><strong>解释：</strong></p>

<p><code>nums</code>&nbsp;中没有数字出现两次。</p>
</div>

<p><strong class="example">示例 3：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>nums = [1,2,2,1]</span></p>

<p><span class="example-io"><b>输出：</b>3</span></p>

<p><strong>解释：</strong></p>

<p>数字 1 和&nbsp;2 出现过两次。<code>1 XOR 2 == 3</code>&nbsp;。</p>
</div>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
<li><code>1 &lt;= nums.length &lt;= 50</code></li>
<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
<li><code>nums</code>&nbsp;中每个数字要么出现过一次，要么出现过两次。</li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一

<!-- tabs:start -->

#### Python3

```python

```

#### Java

```java

```

#### C++

```cpp

```

#### Go

```go

```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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---
comments: true
difficulty: Easy
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3158.Find%20the%20XOR%20of%20Numbers%20Which%20Appear%20Twice/README_EN.md
---

<!-- problem:start -->

# [3158. Find the XOR of Numbers Which Appear Twice](https://leetcode.com/problems/find-the-xor-of-numbers-which-appear-twice)

[中文文档](/solution/3100-3199/3158.Find%20the%20XOR%20of%20Numbers%20Which%20Appear%20Twice/README.md)

## Description

<!-- description:start -->

<p>You are given an array <code>nums</code>, where each number in the array appears <strong>either</strong><em> </em>once<em> </em>or<em> </em>twice.</p>

<p>Return the bitwise<em> </em><code>XOR</code> of all the numbers that appear twice in the array, or 0 if no number appears twice.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,1,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

<p><strong>Explanation:</strong></p>

<p>The only number that appears twice in&nbsp;<code>nums</code>&nbsp;is 1.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>

<p><strong>Explanation:</strong></p>

<p>No number appears twice in&nbsp;<code>nums</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,2,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>Numbers 1 and 2 appeared twice. <code>1 XOR 2 == 3</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
<li><code>1 &lt;= nums.length &lt;= 50</code></li>
<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
<li>Each number in <code>nums</code> appears either once or twice.</li>
</ul>

<!-- description:end -->

## Solutions

<!-- solution:start -->

### Solution 1

<!-- tabs:start -->

#### Python3

```python

```

#### Java

```java

```

#### C++

```cpp

```

#### Go

```go

```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
104 changes: 104 additions & 0 deletions solution/3100-3199/3159.Find Occurrences of an Element in an Array/README.md
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---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3159.Find%20Occurrences%20of%20an%20Element%20in%20an%20Array/README.md
---

<!-- problem:start -->

# [3159. 查询数组中元素的出现位置](https://leetcode.cn/problems/find-occurrences-of-an-element-in-an-array)

[English Version](/solution/3100-3199/3159.Find%20Occurrences%20of%20an%20Element%20in%20an%20Array/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给你一个整数数组&nbsp;<code>nums</code>&nbsp;，一个整数数组&nbsp;<code>queries</code>&nbsp;和一个整数&nbsp;<code>x</code>&nbsp;。</p>

<p>对于每个查询&nbsp;<code>queries[i]</code>&nbsp;，你需要找到&nbsp;<code>nums</code>&nbsp;中第&nbsp;<code>queries[i]</code>&nbsp;个&nbsp;<code>x</code>&nbsp;的位置，并返回它的下标。如果数组中&nbsp;<code>x</code>&nbsp;的出现次数少于&nbsp;<code>queries[i]</code>&nbsp;，该查询的答案为 -1 。</p>

<p>请你返回一个整数数组&nbsp;<code>answer</code>&nbsp;，包含所有查询的答案。</p>

<p>&nbsp;</p>

<p><strong class="example">示例 1：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>nums = [1,3,1,7], queries = [1,3,2,4], x = 1</span></p>

<p><span class="example-io"><b>输出：</b>[0,-1,2,-1]</span></p>

<p><strong>解释：</strong></p>

<ul>
<li>第 1 个查询，第一个 1 出现在下标 0 处。</li>
<li>第 2 个查询，<code>nums</code>&nbsp;中只有两个 1 ，所以答案为 -1 。</li>
<li>第 3 个查询，第二个 1 出现在下标 2 处。</li>
<li>第 4 个查询，<code>nums</code>&nbsp;中只有两个 1 ，所以答案为 -1 。</li>
</ul>
</div>

<p><strong class="example">示例 2：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>nums = [1,2,3], queries = [10], x = 5</span></p>

<p><span class="example-io"><b>输出：</b>[-1]</span></p>

<p><strong>解释：</strong></p>

<ul>
<li>第 1 个查询，<code>nums</code>&nbsp;中没有 5 ，所以答案为 -1 。</li>
</ul>
</div>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
<li><code>1 &lt;= nums.length, queries.length &lt;= 10<sup>5</sup></code></li>
<li><code>1 &lt;= queries[i] &lt;= 10<sup>5</sup></code></li>
<li><code>1 &lt;= nums[i], x &lt;= 10<sup>4</sup></code></li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一

<!-- tabs:start -->

#### Python3

```python

```

#### Java

```java

```

#### C++

```cpp

```

#### Go

```go

```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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