1
0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-03-14 16:22:24 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee
leetcode-problemset/leetcode-cn/problem (Chinese)/长度为 K 的子数组的能量值 II [find-the-power-of-k-size-subarrays-ii].html

61 lines
2.3 KiB
HTML
Raw Normal View History

2024-08-27 23:06:51 +08:00
<p>给你一个长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;和一个正整数&nbsp;<code>k</code>&nbsp;</p>
<p>一个数组的 <strong>能量值</strong> 定义为:</p>
<ul>
2025-01-09 20:29:41 +08:00
<li>如果 <strong>所有</strong>&nbsp;元素都是依次&nbsp;<strong>连续</strong>(即 <code>nums[i] + 1 = nums[i + 1]</code><code>i &lt; n</code>)且 <strong>上升</strong> 的,那么能量值为 <strong>最大</strong>&nbsp;的元素。</li>
2024-08-27 23:06:51 +08:00
<li>否则为 -1 。</li>
</ul>
<p>你需要求出 <code>nums</code>&nbsp;中所有长度为 <code>k</code>&nbsp;&nbsp;<span data-keyword="subarray-nonempty">子数组</span>&nbsp;的能量值。</p>
<p>请你返回一个长度为 <code>n - k + 1</code>&nbsp;的整数数组&nbsp;<code>results</code>&nbsp;,其中&nbsp;<code>results[i]</code>&nbsp;是子数组&nbsp;<code>nums[i..(i + k - 1)]</code>&nbsp;的能量值。</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 = [1,2,3,4,3,2,5], k = 3</span></p>
<p><b>输出:</b>[3,4,-1,-1,-1]</p>
<p><strong>解释:</strong></p>
<p><code>nums</code>&nbsp;中总共有 5 个长度为 3 的子数组:</p>
<ul>
<li><code>[1, 2, 3]</code>&nbsp;中最大元素为 3 。</li>
<li><code>[2, 3, 4]</code>&nbsp;中最大元素为 4 。</li>
<li><code>[3, 4, 3]</code>&nbsp;中元素 <strong>不是</strong>&nbsp;连续的。</li>
<li><code>[4, 3, 2]</code>&nbsp;中元素 <b>不是</b>&nbsp;上升的。</li>
<li><code>[3, 2, 5]</code>&nbsp;中元素 <strong>不是</strong>&nbsp;连续的。</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 = [2,2,2,2,2], k = 4</span></p>
<p><span class="example-io"><b>输出:</b>[-1,-1]</span></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 = [3,2,3,2,3,2], k = 2</span></p>
<p><span class="example-io"><b>输出:</b>[-1,3,-1,3,-1]</span></p>
</div>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
<li><code>1 &lt;= k &lt;= n</code></li>
</ul>