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leetcode-cn/problem (Chinese)/数组中两个数的最大异或值 [maximum-xor-of-two-numbers-in-an-array].html
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<p>给你一个整数数组 <code>nums</code> ，返回<em> </em><code>nums[i] XOR nums[j]</code> 的最大运算结果，其中 <code>0 ≤ i ≤ j < n</code> 。</p>
<p>给你一个整数数组 <code>nums</code> ，返回<em> </em><code>nums[i] XOR nums[j]</code> 的最大运算结果，其中 <code>0 ≤ i ≤ j &lt; n</code> 。</p>
<p><strong>进阶：</strong>你可以在 <code>O(n)</code> 的时间解决这个问题吗？</p>
<p> </p>
<p>&nbsp;</p>
<div class="original__bRMd">
<div>
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<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>nums = [0]
<strong>输出：</strong>0
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>nums = [2,4]
<strong>输出：</strong>6
</pre>
<p><strong>示例 4：</strong></p>
<pre>
<strong>输入：</strong>nums = [8,10,2]
<strong>输出：</strong>10
</pre>
<p><strong>示例 5：</strong></p>
<pre>
<strong>输入：</strong>nums = [14,70,53,83,49,91,36,80,92,51,66,70]
<strong>输出：</strong>127
</pre>
<p> </p>
<p>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= nums.length <= 2 * 10<sup>4</sup></code></li>
<li><code>0 <= nums[i] <= 2<sup>31</sup> - 1</code></li>
<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
<li><code>0 &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
</div>
</div>