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.

题目没有对空间的要求，因此最简单的方法是新建两个链表，一个存小的，一个存大的。下面的即是这种方法。
当然也可以采用双指针的方法，一个指向小的，一个指向大的，遍历数组，然后小的挪到一块，大的挪到一块，只不过会出现很多指针的连接操作。

/**
 * 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 || head.next==null) {  
            return head;  
        }  
        ListNode s = new ListNode(0);//small partition  
        ListNode p = s;  
        ListNode l = new ListNode(0);//large partition  
        ListNode q = l;  
        while(head!=null){  
            if (head.val<x) {  
                p.next = head;  
                p = p.next;  
            }else {  
                q.next = head;  
                q = q.next;  
            }  
            head = head.next;  
        }  
        q.next = null;  
        p.next = l.next;  
        return