原题链接在这里：

题目：

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. 

题解:

需要建一个dunmy，最后return dunmy.next. 如此可以避免边界问题的讨论，例如头两个点的转换。

Time Complexity: O(n), n是list的长度. Space: O(1).

AC Java:

1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { val = x; } 7  * } 8  */ 9 public class Solution {10     public ListNode swapPairs(ListNode head) {11         if(head == null || head.next == null){12             return head;13         }14         15         ListNode dummy = new ListNode(0);16         dummy.next = head;17         18         ListNode mark = dummy;19         ListNode A;20         ListNode B;21         while(mark.next != null && mark.next.next != null){22             A = mark.next;23             B = mark.next.next;24             A.next = B.next;25             B.next = A;26             mark.next = B;27             mark = mark.next.next;28         }29         return dummy.next;30     }31 }

Recursion解法.

Time Complexity: O(n). Space: O(n), stack space.

AC Java: 

1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { val = x; } 7  * } 8  */ 9 public class Solution {10     public ListNode swapPairs(ListNode head) {11         if(head == null || head.next == null){12             return head;13         }14         ListNode n = head.next;15         head.next = swapPairs(n.next);16         n.next = head;17         return n;18     }19 }

跟上.