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Description
tab: English
<p>Given the <code>head</code> of a linked list and a value <code>x</code>, partition it such that all nodes <strong>less than</strong> <code>x</code> come before nodes <strong>greater than or equal</strong> to <code>x</code>.</p>
<p>You should <strong>preserve</strong> the original relative order of the nodes in each of the two partitions.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/partition.jpg" style="width: 662px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,4,3,2,5,2], x = 3
<strong>Output:</strong> [1,2,2,4,3,5]
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> head = [2,1], x = 2
<strong>Output:</strong> [1,2]
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li>The number of nodes in the list is in the range <code>[0, 200]</code>.</li>
<li><code>-100 <= Node.val <= 100</code></li>
<li><code>-200 <= x <= 200</code></li>
</ul>
---
[submissions](https://leetcode.com/problems/partition-list/submissions/) | [solutions](https://leetcode.com/problems/partition-list/solutions/)
tab: 中文
<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>
<p>你应当 <strong>保留</strong> 两个分区中每个节点的初始相对位置。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/partition.jpg" style="width: 662px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,4,3,2,5,2], x = 3
<strong>输出</strong>：[1,2,2,4,3,5]
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>head = [2,1], x = 2
<strong>输出</strong>：[1,2]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>链表中节点的数目在范围 <code>[0, 200]</code> 内</li>
<li><code>-100 <= Node.val <= 100</code></li>
<li><code>-200 <= x <= 200</code></li>
</ul>
---
[提交记录](https://leetcode.cn/problems/partition-list/submissions/) | [题解](https://leetcode.cn/problems/partition-list/solution/)
Solutions & Notes
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