1.题目解析

2.讲解算法原理 

• 让前两个节点后面的链表先逆置，并且把头结点返回
• 把当前两个结点进行交换，并且找到新的头结点
• 把后面链表直接添加到当前已经逆置的两个节点后面 

3.编写代码

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