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

You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example:

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

本题的解题思路很简单。我们首先提取为list添加一个头结点。然后我设置三个指针,pre,cur,post。cur和post是需要两两交换的指针。令pre.next = post,然后post.next = cur。这样子就达到了交换的目的，交换后继续更新三个指针的位置。直到最后面得出结果。

/**
 * 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 || null == head.next){
            return head;
        }
        ListNode first = new ListNode(0);
        first.next = head;
        ListNode pre = first;
        ListNode cur = head;
        ListNode post = cur.next;
        while(cur != null && post != cur){
            ListNode newNode = post.next;
            pre.next = post;
            post.next = cur;
            
            cur.next = newNode;
            pre = cur;
            cur = pre.next;
            if(cur == null){
                return first.next;
            } else if(cur.next != null){
                post = cur.next;
            } else {
                post = cur;
            }
        }
        return first.next;   
    }
}

使用递归进行操作相对于迭代而言会比较抽象一点，但是也不是很难理解。其核心思想是和迭代的一样的。每次两对两对的更新。只不过递归是从后往前交换，而迭代时从前往后交换。

class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode nextnextSwapped = swapPairs(head.next.next);
        ListNode newHead = head.next;
        newHead.next = head;
        head.next = nextnextSwapped;
        return newHead;    
    }
}