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.

1         public static LinkedListNode SwapNodesinPairs2(LinkedListNode head) 2         { 3             if (head == null || head.Next == null) 4                 return head; 5  6             LinkedListNode fakehead = new LinkedListNode(); 7             fakehead.Next = head; 8  9             LinkedListNode l1 = head;10             LinkedListNode l2 = head;11             LinkedListNode prev = fakehead;12             LinkedListNode ret = head.Next;13             LinkedListNode next = head;14             15             while (next != null && next.Next != null)16             {17                 l1 = next;18                 l2 = next.Next;19                 next = next.Next.Next;20                 l2.Next = l1;21                 l1.Next = next;22                 prev.Next = l2;23                 prev = l1;24             }25 26             return ret;27         }

代码分析：

　　看来我的总结是正确的，每遇到Linked List的题都在前面加个fakehead. 后面的逻辑就方便很多，不用判断这个那个。