Swap Nodes in Pairs

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.

这道题如果直接交换值是很easy的

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

按照题目要求不交换值，交换节点的引用

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode swapPairs(ListNode head) {
14         if(null == head || null == head.next)
15             return head;
16         ListNode behind = head;
17         ListNode front = behind.next;
18         ListNode temp;
19         ListNode pre;
20         //交换开头两个节点
21         
22         behind.next = front.next;
23         front.next = behind;
24         
25         head = front;
26       //交换一下front和behind
27         temp = behind;
28         behind = front;
29         front = temp;
30         
31         pre = front;//倒数第二个指针
32         
33         while(null != front.next && null != front.next.next){
34             behind = front.next;
35             front = behind.next;
36             
37             behind.next = front.next;
38             front.next = behind;
39             
40             temp = behind;
41             behind = front;
42             front = temp;
43             
44             pre.next = behind;
45             pre = front;
46         }
47         
48         return head;
49     }
50 }