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 tox.

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.

//使用两个队列，一个存储小于x的，一个存储大于等于x的，然后连接两个队列

class Solution {

    public ListNode partition(ListNode head, int x) {

        ListNode dummy1 = new ListNode(0);

        ListNode dummy2 = new ListNode(0);

        ListNode curr1 = dummy1;

        ListNode curr2 = dummy2;

        while (head != null) {

            if (head.val < x) {

                curr1.next = head;

                curr1 = head;

            } else {

                curr2.next = head;

                curr2 = head;

            }

            head = head.next;

        }

        

        curr2.next = null;//一定要有这句话，比如 5->6->1->2, x=3，c1：1->2,c2：5->6，我们让2指向5了，如果还有这句话，会保持6->1，会形成个cycle

        curr1.next = dummy2.next;

        return dummy1.next;

    }

}