# Partition List

**Problem Statment**

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

**Example**

Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5

**Solution**

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode partition(ListNode head, int x) { if (head == null) { return head; } ListNode smallHead = new ListNode(0); ListNode small = smallHead; ListNode largereqHead = new ListNode(0); ListNode largereq = largereqHead; ListNode current = head; while (current != null) { if (current.val < x) { small.next = current; small = small.next; } else { largereq.next = current; largereq = largereq.next; } current = current.next; } largereq.next = null; small.next = largereqHead.next; return smallHead.next; } }