Given a linked list, swap every two adjacent nodes and return its head.

For example,

Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

Tips：给定一个链表，在不改变链表的值，只改变结点的指针的情况下完成相邻两个结点的交换。

如：Input：1->2->3->4

　　Output：2->1->4->3

思路：设置两个指针，first second，分别为当前指针的next以及next.next。

将 first.next指向second.next;（1->3）

然后将second.next指向first(2->1)。

再将当前指针后移：cur.next=second; cur = first;

package medium;import dataStructure.ListNode;public class L24SwapNodesInPairs {    public ListNode swapPairs(ListNode head) {        if (head == null)            return head;        ListNode node = new ListNode(0);        node.next = head;        ListNode cur = node;                while (cur.next != null && cur.next.next!=null) {            ListNode first = cur.next;            ListNode second = cur.next.next;            first.next = second.next;//1->3            second.next=first;//2->1->3            cur.next=second;            cur = first;        }        return node.next;    }    public static void main(String[] args) {        ListNode head1 = new ListNode(1);        ListNode head2 = new ListNode(2);        ListNode head3 = new ListNode(3);        ListNode head4 = new ListNode(4);        ListNode head5 = new ListNode(5);        ListNode head6 = new ListNode(6);        ListNode head7 = null;        head1.next = head2;        head2.next = head3;        head3.next = head4;        head4.next = head5;        head5.next = head6;        head6.next = head7;        L24SwapNodesInPairs l24 = new L24SwapNodesInPairs();        ListNode head = l24.swapPairs(head1);        while (head != null ) {            System.out.println(head.val);            head=head.next;        }    }}