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solution/2600-2699/2674.Split a Circular Linked List/README.md
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<p><strong>提示:</strong></p>
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<ul>
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- <li><code>list</code> 中的节点数范围为 <code>[2, 105]</code></li>
+ <li><code>list</code> 中的节点数范围为 <code>[2, 10<sup>5</sup>]</code></li>
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<li><code>0 <= Node.val <= 10<sup>9</sup></code></li>
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<li><code>LastNode.next = FirstNode</code> ,其中 <code>LastNode</code> 是链表的最后一个节点,<code>FirstNode</code> 是第一个节点。</li>
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</ul>
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