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.

题目的意思是将小于x的元素放在list的前面，大于等于的放在后面，其中不能打乱原来

list中元素的顺序。

我的想法是用两个queue分别存储小于x和大于等于x的元素，再分别出队列组成新的list

就是要求的list。

其实queue就是实现的linkedlist接口。九章算法直接新建两个链表分别连接小于x和大于

等于x的元素，再将小于的链表指向大于的链表，形成最后的list。

public ListNode partition(ListNode head, int x) {if (head == null) return null;ListNode leftDummy = new ListNode(0);ListNode rightDummy = new ListNode(0);ListNode left = leftDummy, right = rightDummy;while (head != null) {if (head.val < x) {left.next = head;left = head;} else {right.next = head;right = head;}head = head.next;}right.next = null;left.next = rightDummy.next;return leftDummy.next;}