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feat: add new lc problems: No.2089+
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# [2089. 找出数组排序后的目标下标](https://leetcode-cn.com/problems/find-target-indices-after-sorting-array)
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[English Version](/solution/2000-2099/2089.Find%20Target%20Indices%20After%20Sorting%20Array/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 以及一个目标元素 <code>target</code> 。</p>
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<p><strong>目标下标</strong> 是一个满足&nbsp;<code>nums[i] == target</code> 的下标 <code>i</code> 。</p>
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<p>将 <code>nums</code> 按 <strong>非递减</strong> 顺序排序后,返回由 <code>nums</code> 中目标下标组成的列表。如果不存在目标下标,返回一个 <strong>空</strong> 列表。返回的列表必须按 <strong>递增</strong> 顺序排列。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>nums = [1,2,5,2,3], target = 2
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<strong>输出:</strong>[1,2]
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<strong>解释:</strong>排序后,nums 变为 [1,<em><strong>2</strong></em>,<em><strong>2</strong></em>,3,5] 。
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满足 nums[i] == 2 的下标是 1 和 2 。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>nums = [1,2,5,2,3], target = 3
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<strong>输出:</strong>[3]
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<strong>解释:</strong>排序后,nums 变为 [1,2,2,<em><strong>3</strong></em>,5] 。
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满足 nums[i] == 3 的下标是 3 。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入:</strong>nums = [1,2,5,2,3], target = 5
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<strong>输出:</strong>[4]
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<strong>解释:</strong>排序后,nums 变为 [1,2,2,3,<em><strong>5</strong></em>] 。
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满足 nums[i] == 5 的下标是 4 。
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre><strong>输入:</strong>nums = [1,2,5,2,3], target = 4
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<strong>输出:</strong>[]
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<strong>解释:</strong>nums 中不含值为 4 的元素。
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</pre>
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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;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i], target &lt;= 100</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2089. Find Target Indices After Sorting Array](https://leetcode.com/problems/find-target-indices-after-sorting-array)
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[中文文档](/solution/2000-2099/2089.Find%20Target%20Indices%20After%20Sorting%20Array/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and a target element <code>target</code>.</p>
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<p>A <strong>target index</strong> is an index <code>i</code> such that <code>nums[i] == target</code>.</p>
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<p>Return <em>a list of the target indices of</em> <code>nums</code> after<em> sorting </em><code>nums</code><em> in <strong>non-decreasing</strong> order</em>. If there are no target indices, return <em>an <strong>empty</strong> list</em>. The returned list must be sorted in <strong>increasing</strong> order.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,5,2,3], target = 2
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<strong>Output:</strong> [1,2]
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<strong>Explanation:</strong> After sorting, nums is [1,<u><strong>2</strong></u>,<u><strong>2</strong></u>,3,5].
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The indices where nums[i] == 2 are 1 and 2.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,5,2,3], target = 3
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<strong>Output:</strong> [3]
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<strong>Explanation:</strong> After sorting, nums is [1,2,2,<u><strong>3</strong></u>,5].
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The index where nums[i] == 3 is 3.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,5,2,3], target = 5
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<strong>Output:</strong> [4]
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<strong>Explanation:</strong> After sorting, nums is [1,2,2,3,<u><strong>5</strong></u>].
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The index where nums[i] == 5 is 4.
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,5,2,3], target = 4
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<strong>Output:</strong> []
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<strong>Explanation:</strong> There are no elements in nums with value 4.
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</pre>
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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;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i], target &lt;= 100</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2090. 半径为 k 的子数组平均值](https://leetcode-cn.com/problems/k-radius-subarray-averages)
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[English Version](/solution/2000-2099/2090.K%20Radius%20Subarray%20Averages/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个下标从 <strong>0</strong> 开始的数组 <code>nums</code> ,数组中有 <code>n</code> 个整数,另给你一个整数 <code>k</code> 。</p>
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<p><strong>半径为 k 的子数组平均值</strong> 是指:<code>nums</code> 中一个以下标 <code>i</code> 为 <strong>中心</strong> 且 <strong>半径</strong> 为 <code>k</code> 的子数组中所有元素的平均值,即下标在&nbsp;<code>i - k</code> 和 <code>i + k</code> 范围(<strong>含</strong> <code>i - k</code> 和 <code>i + k</code>)内所有元素的平均值。如果在下标 <code>i</code> 前或后不足 <code>k</code> 个元素,那么<strong> 半径为 k 的子数组平均值 </strong>是 <code>-1</code> 。</p>
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<p>构建并返回一个长度为 <code>n</code> 的数组<em> </em><code>avgs</code><em> </em>,其中<em> </em><code>avgs[i]</code><em> </em>是以下标 <code>i</code> 为中心的子数组的<strong> 半径为 k 的子数组平均值 </strong>。</p>
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<p><code>x</code> 个元素的 <strong>平均值</strong> 是 <code>x</code> 个元素相加之和除以 <code>x</code> ,此时使用截断式 <strong>整数除法</strong> ,即需要去掉结果的小数部分。</p>
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<ul>
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<li>例如,四个元素 <code>2</code>、<code>3</code>、<code>1</code> 和 <code>5</code> 的平均值是 <code>(2 + 3 + 1 + 5) / 4 = 11 / 4 = 3.75</code>,截断后得到 <code>3</code> 。</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2090.K%20Radius%20Subarray%20Averages/images/eg1.png" style="width: 343px; height: 119px;" /></p>
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<pre>
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<strong>输入:</strong>nums = [7,4,3,9,1,8,5,2,6], k = 3
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<strong>输出:</strong>[-1,-1,-1,5,4,4,-1,-1,-1]
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<strong>解释:</strong>
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- avg[0]、avg[1] 和 avg[2] 是 -1 ,因为在这几个下标前的元素数量都不足 k 个。
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- 中心为下标 3 且半径为 3 的子数组的元素总和是:7 +たす 4 +たす 3 +たす 9 +たす 1 +たす 8 +たす 5 =わ 37 。
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使用截断式 <strong>整数除法</strong>,avg[3] = 37 / 7 = 5 。
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- 中心为下标 4 的子数组,avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4 。
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- 中心为下标 5 的子数组,avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4 。
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- avg[6]、avg[7] 和 avg[8] 是 -1 ,因为在这几个下标后的元素数量都不足 k 个。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [100000], k = 0
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<strong>输出:</strong>[100000]
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<strong>解释:</strong>
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- 中心为下标 0 且半径 0 的子数组的元素总和是:100000 。
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avg[0] = 100000 / 1 = 100000 。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [8], k = 100000
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<strong>输出:</strong>[-1]
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<strong>解释:</strong>
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- avg[0] 是 -1 ,因为在下标 0 前后的元素数量均不足 k 。
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</pre>
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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 == nums.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>0 &lt;= nums[i], k &lt;= 10<sup>5</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2090. K Radius Subarray Averages](https://leetcode.com/problems/k-radius-subarray-averages)
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[中文文档](/solution/2000-2099/2090.K%20Radius%20Subarray%20Averages/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of <code>n</code> integers, and an integer <code>k</code>.</p>
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<p>The <strong>k-radius average</strong> for a subarray of <code>nums</code> <strong>centered</strong> at some index <code>i</code> with the <strong>radius</strong> <code>k</code> is the average of <strong>all</strong> elements in <code>nums</code> between the indices <code>i - k</code> and <code>i + k</code> (<strong>inclusive</strong>). If there are less than <code>k</code> elements before <strong>or</strong> after the index <code>i</code>, then the <strong>k-radius average</strong> is <code>-1</code>.</p>
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<p>Build and return <em>an array </em><code>avgs</code><em> of length </em><code>n</code><em> where </em><code>avgs[i]</code><em> is the <strong>k-radius average</strong> for the subarray centered at index </em><code>i</code>.</p>
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<p>The <strong>average</strong> of <code>x</code> elements is the sum of the <code>x</code> elements divided by <code>x</code>, using <strong>integer division</strong>. The integer division truncates toward zero, which means losing its fractional part.</p>
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<ul>
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<li>For example, the average of four elements <code>2</code>, <code>3</code>, <code>1</code>, and <code>5</code> is <code>(2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75</code>, which truncates to <code>2</code>.</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2090.K%20Radius%20Subarray%20Averages/images/eg1.png" style="width: 343px; height: 119px;" />
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<pre>
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<strong>Input:</strong> nums = [7,4,3,9,1,8,5,2,6], k = 3
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<strong>Output:</strong> [-1,-1,-1,5,4,4,-1,-1,-1]
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<strong>Explanation:</strong>
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- avg[0], avg[1], and avg[2] are -1 because there are less than k elements <strong>before</strong> each index.
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- The sum of the subarray centered at index 3 with radius 3 is: 7 +たす 4 +たす 3 +たす 9 +たす 1 +たす 8 +たす 5 =わ 37.
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Using <strong>integer division</strong>, avg[3] = 37 / 7 = 5.
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- For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.
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- For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.
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- avg[6], avg[7], and avg[8] are -1 because there are less than k elements <strong>after</strong> each index.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [100000], k = 0
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<strong>Output:</strong> [100000]
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<strong>Explanation:</strong>
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- The sum of the subarray centered at index 0 with radius 0 is: 100000.
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avg[0] = 100000 / 1 = 100000.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [8], k = 100000
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<strong>Output:</strong> [-1]
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<strong>Explanation:</strong>
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- avg[0] is -1 because there are less than k elements before and after index 0.
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</pre>
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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>n == nums.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>0 &lt;= nums[i], k &lt;= 10<sup>5</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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