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.

准备左右两个链表，左链表放比x小的节点，右链表放大于等于x的节点，最后两个链表再进行连接。

	public ListNode partition(ListNode head, int x) {
		ListNode left = new ListNode(-1), tmpL = left;
		ListNode right = new ListNode(-1), tmpR = right;
		while (head != null) {
			if (head.val < x) {
				left.next = head;
				left = left.next;
			} else {
				right.next = head;
				right = right.next;
			}
			head = head.next;
		}
		left.next = tmpR.next;
		right.next = null;
		return tmpL.next;
	}