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leetcode-problemset/leetcode-cn/problem (Chinese)/数组元素相等的最小操作次数 [minimum-operations-to-equalize-array].html
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<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code></p>
<p>在一次操作中,可以选择任意子数组 <code>nums[l...r]</code> <code>0 &lt;= l &lt;= r &lt; n</code>),并将该子数组中的每个元素&nbsp;<strong>替换&nbsp;</strong>为所有元素的&nbsp;<strong>按位与bitwise AND</strong>结果。</p>
<p>返回使数组 <code>nums</code> 中所有元素相等所需的最小操作次数。</p>
<p><strong>子数组&nbsp;</strong>是数组中连续的、非空的元素序列。</p>
<p>&nbsp;</p>
<p><strong class="example">示例 1</strong></p>
<div class="example-block">
<p><strong>输入:</strong> <span class="example-io">nums = [1,2]</span></p>
<p><strong>输出:</strong> <span class="example-io">1</span></p>
<p><strong>解释:</strong></p>
<p>选择 <code>nums[0...1]</code><code>(1 AND 2) = 0</code>,因此数组变为 <code>[0, 0]</code>,所有元素在一次操作后相等。</p>
</div>
<p><strong class="example">示例 2</strong></p>
<div class="example-block">
<p><strong>输入:</strong> <span class="example-io">nums = [5,5,5]</span></p>
<p><strong>输出:</strong> <span class="example-io">0</span></p>
<p><strong>解释:</strong></p>
<p><code>nums</code> 本身是 <code>[5, 5, 5]</code>,所有元素已经相等,因此不需要任何操作。</p>
</div>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 100</code></li>
<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>