Leetcode 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.

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    // this solution doesn't pass OJ but I think it actually solves the problem.
    public ListNode PartitionInPlace(ListNode head, int x) {
        if (head == null) return null;
        
        ListNode p1 = head, p2 = head;
        
        while (p2 != null)
        {
            if (p2.val < x)
            {
                if (p1 != p2)
                {
                    var tmp = p1.val;
                    p1.val = p2.val;
                    p2.val = tmp;
                }
                
                p1 = p1.next;
            }
            
            p2 = p2.next;
        }
        
        return head;        
    }
    
    public ListNode Partition(ListNode head, int x) {
        if (head == null) return null;
        
        ListNode p1 = new ListNode(1), p2 = new ListNode(2), cur1 = p1, cur2 = p2, cur = head;
        
        while (cur != null)
        {
            var tmp = cur.next;
            cur.next = null;
            if (cur.val < x)
            {
                cur1.next = cur;
                cur1 = cur1.next;
            }
            else
            {
                cur2.next = cur;
                cur2 = cur2.next;
            }
            cur = tmp;
        }
        
        cur1.next = p2.next;
        return p1.next;        
    }
}