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

注意：
分开两个大小链表来连接大小的节点，节点连接到大小链表之后，不需要将其的next 更新为 null，到最后设置 big 最后一个的 next 为 null 即可。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param x: an integer
     * @return: a ListNode 
     */
    public ListNode partition(ListNode head, int x) {
        // write your code here
        ListNode dummy1 = new ListNode(0);
        ListNode smallLast = dummy1;
        ListNode dummy2 = new ListNode(0);
        ListNode bigLast = dummy2;
        while (head != null) {
            if (head.val < x) {
                smallLast.next = head;
                smallLast = smallLast.next;
            } else {
                bigLast.next = head;
                bigLast = bigLast.next;
            }
            head = head.next;
        }
        smallLast.next = dummy2.next;
        bigLast.next = null;
        return dummy1.next;
    }
}