Partition List，拆分链表

问题描述：

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.

算法分析：开始我只是想在原来链表上进行操作，但是无法返回头结点。这道题区别链表反转，链表反转不用新建链表，只用在原有的链表上操作就行了。这道题，要新建两个新的链表，一个链表的元素全部小于目标值，另一个链表的元素全部大于目标值。然后把这两个链表连接起来。

public ListNode partition(ListNode head, int x)
	{
		if(head == null || head.next == null)
		{
			return head;
		}
		ListNode lessHead = new ListNode(0);
		ListNode greaterHead = new ListNode(0);
		ListNode less = lessHead, greater = greaterHead;
		ListNode node = head;
		while(node != null)
		{
			ListNode temp = node.next;
			if(node.val < x)
			{
				less.next = node;
				less = less.next;
				less.next = null;
			}
			else
			{
				greater.next = node;
				greater = greater.next;
				greater.next = null;
			}
			node = temp;
		}
		less.next = greaterHead.next;
		return lessHead.next;
	}