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

Example:

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

Note:

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

 

要求把相邻的2个节点两两互换，还必须是换指针而不能是只换值

这里我们用递归的方法来处理，两个指针l1和l2，l1-->l2，我们先把l1和l2换了，然后对l1.next.next继续相同的方法递归下去

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode swapPairs(ListNode head) {        if (head==null || head.next==null) return head;        ListNode res = head.next;        head.next = swapPairs(head.next.next);        res.next = head;        return res;    }}