<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>&nbsp;</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>&nbsp;</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 &lt;= Node.val &lt;= 100</code></li>
	<li><code>-200 &lt;= x &lt;= 200</code></li>
</ul>

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode* d1 = new ListNode();
        ListNode* d2 = new ListNode();
        ListNode* t1 = d1;
        ListNode* t2 = d2;
        while (head) {
            if (head->val < x) {
                t1->next = head;
                t1 = t1->next;
            } else {
                t2->next = head;
                t2 = t2->next;
            }
            head = head->next;
        }
        t1->next = d2->next;
        t2->next = nullptr;
        return d1->next;
    }
};
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