86. Partition List

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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

public ListNode Partition(ListNode head, int x) {
        if(head == null) return head;
        ListNode smallStart =new ListNode(0);
        ListNode largeStart =new ListNode(0);
        ListNode p1 = smallStart;
        ListNode p2 = largeStart;
        //assign small and large pointer
        
        while(head != null)
        {
            if(head.val < x)
            {
                 smallStart.next =head;
                 smallStart = smallStart.next;    
 
            }
            else
            {
                largeStart.next = head;
                largeStart = largeStart.next;
            }
            head = head.next;
        }
        largeStart.next = null;
        smallStart.next = p2.next;
        return p1.next;
    }