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97 lines
3.4 KiB
HTML
97 lines
3.4 KiB
HTML
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<p>给你一个由 <code>n</code> 个整数组成的数组 <code>nums</code>，以及一个大小为 <code>q</code> 的二维整数数组 <code>queries</code>，其中 <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>。</p>
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<p>对于每一个查询，你需要找出 <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code> 中任意 <span data-keyword="subarray">子数组</span> 的 <strong>最大异或值</strong>。</p>
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<p><strong>数组的异或值 </strong>需要对数组 <code>a</code> 反复执行以下操作，直到只剩一个元素，剩下的那个元素就是 <strong>异或值</strong>：</p>
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<ul>
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<li><span class="text-only" data-eleid="9" style="white-space: pre;">对于除最后一个下标以外的所有下标</span> <code>i</code>，同时将 <code>a[i]</code> 替换为 <code>a[i] XOR a[i + 1]</code> 。</li>
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<li>移除数组的最后一个元素。</li>
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</ul>
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<p>返回一个大小为 <code>q</code> 的数组 <code>answer</code>，其中 <code>answer[i]</code> 表示查询 <code>i</code> 的答案。</p>
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<p> </p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">nums = [2,8,4,32,16,1], queries = [[0,2],[1,4],[0,5]]</span></p>
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<p><strong>输出：</strong> <span class="example-io">[12,60,60]</span></p>
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<p><strong>解释：</strong></p>
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<p>在第一个查询中，<code>nums[0..2]</code> 的子数组分别是 <code>[2]</code>, <code>[8]</code>, <code>[4]</code>, <code>[2, 8]</code>, <code>[8, 4]</code>, 和 <code>[2, 8, 4]</code>，它们的异或值分别为 2, 8, 4, 10, 12, 和 6。查询的答案是 12，所有异或值中的最大值。</p>
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<p>在第二个查询中，<code>nums[1..4]</code> 的子数组中最大的异或值是子数组 <code>nums[1..4]</code> 的异或值，为 60。</p>
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<p>在第三个查询中，<code>nums[0..5]</code> 的子数组中最大的异或值是子数组 <code>nums[1..4]</code> 的异或值，为 60。</p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">nums = [0,7,3,2,8,5,1], queries = [[0,3],[1,5],[2,4],[2,6],[5,6]]</span></p>
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<p><strong>输出：</strong> <span class="example-io">[7,14,11,14,5]</span></p>
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<p><strong>解释：</strong></p>
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<table height="70" width="472">
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<thead>
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<tr>
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<th>下标</th>
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<th>nums[l<sub>i</sub>..r<sub>i</sub>]</th>
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<th>最大异或值子数组</th>
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<th>子数组最大异或值</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td>0</td>
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<td>[0, 7, 3, 2]</td>
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<td>[7]</td>
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<td>7</td>
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</tr>
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<tr>
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<td>1</td>
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<td>[7, 3, 2, 8, 5]</td>
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<td>[7, 3, 2, 8]</td>
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<td>14</td>
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</tr>
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<tr>
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<td>2</td>
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<td>[3, 2, 8]</td>
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<td>[3, 2, 8]</td>
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<td>11</td>
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</tr>
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<tr>
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<td>3</td>
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<td>[3, 2, 8, 5, 1]</td>
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<td>[2, 8, 5, 1]</td>
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<td>14</td>
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</tr>
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<tr>
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<td>4</td>
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<td>[5, 1]</td>
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<td>[5]</td>
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<td>5</td>
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</tr>
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</tbody>
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</table>
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</div>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= n == nums.length <= 2000</code></li>
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<li><code>0 <= nums[i] <= 2<sup>31</sup> - 1</code></li>
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<li><code>1 <= q == queries.length <= 10<sup>5</sup></code></li>
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<li><code>queries[i].length == 2</code></li>
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<li><code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code></li>
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<li><code>0 <= l<sub>i</sub> <= r<sub>i</sub> <= n - 1</code></li>
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</ul>
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