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小墨 authored 2023-12-09 18:42 +08:00 . 存量题库数据更新
<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of <strong>distinct</strong> integers. You want to rearrange the elements in the array such that every element in the rearranged array is <strong>not</strong> equal to the <strong>average</strong> of its neighbors.</p>
<p>More formally, the rearranged array should have the property such that for every <code>i</code> in the range <code>1 &lt;= i &lt; nums.length - 1</code>, <code>(nums[i-1] + nums[i+1]) / 2</code> is <strong>not</strong> equal to <code>nums[i]</code>.</p>
<p>Return <em><strong>any</strong> rearrangement of </em><code>nums</code><em> that meets the requirements</em>.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> [1,2,4,5,3]
<strong>Explanation:</strong>
When i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5.
When i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5.
When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [6,2,0,9,7]
<strong>Output:</strong> [9,7,6,2,0]
<strong>Explanation:</strong>
When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
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力扣题库(完整版)
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