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"translatedContent": "<p>给你一个链表的头节点 <code>head</code> 和一个特定值<em> </em><code>x</code> ,请你对链表进行分隔,使得所有 <strong>小于</strong> <code>x</code> 的节点都出现在 <strong>大于或等于</strong> <code>x</code> 的节点之前。</p>\n\n<p>你不需要 <strong>保留</strong> 每个分区中各节点的初始相对位置。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/04/partition.jpg\" style=\"width: 662px; height: 222px;\" />\n<pre>\n<strong>输入:</strong>head = [1,4,3,2,5,2], x = 3\n<strong>输出</strong>:[1,2,2,4,3,5]\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>head = [2,1], x = 2\n<strong>输出</strong>:[1,2]\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>链表中节点的数目在范围 <code>[0, 200]</code> 内</li>\n\t<li><code>-100 <= Node.val <= 100</code></li>\n\t<li><code>-200 <= x <= 200</code></li>\n</ul>\n",
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"difficulty": "Medium",
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@@ -143,7 +143,7 @@
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"hints": [
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"这个问题有很多解法,其中大部分都有最优的运行时间。有些代码比其他代码更短,更干净。你可以想出不同的解法吗?",
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"考虑元素不必保持相同的相对顺序。我们只需要确保小于基准点的元素必须位于比基准点大的元素之前。这有助于你想出更多的解法吗?"
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