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 {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
	public ListNode partition(ListNode head, int x) {
		ListNode lin = head;
		ListNode small = new ListNode(0);
		ListNode hs = small;
		ListNode large = new ListNode(0);
		ListNode hl = large;
		while(lin!=null){
			ListNode ln = new ListNode(lin.val);
			if(lin.val<x){
				small.next = ln;
				small = small.next;
			}else{
				large.next = ln;
				large = large.next;
			}
			lin = lin.next; 
		}
		small.next = hl.next;
		return hs.next;
	}
}

[LeetCode]Partition List