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.

 

 1 public class Solution {

 2     public ListNode swapPairs(ListNode head) {

 3         // Note: The Solution object is instantiated only once and is reused by each test case.

 4         if(head == null || head.next == null) return head;

 5         ListNode newhead = head.next;

 6         ListNode first = head;

 7         ListNode second = head;

 8         ListNode pos = head;

 9         while(pos != null && pos.next != null){

10             first = pos;

11             second = pos.next;

12             pos = second.next;

13             if(pos == null || pos.next == null)

14                 first.next = pos;

15             else

16                 first.next = pos.next;

17             second.next = first;

18         }

19         return newhead;

20     }

21 }

 

 

 1  public class Solution {

 2     public ListNode swapPairs(ListNode head) {

 3         // Start typing your Java solution below

 4         // DO NOT write main() function

 5         ListNode helper = new ListNode(0);

 6         helper.next = head;

 7         ListNode n1 = helper, n2=head;

 8         

 9         while(n2!=null && n2.next!=null){

10             ListNode temp = n2.next.next;

11             n2.next.next=n1.next;

12             n1.next=n2.next;

13             n2.next=temp;

14             n1=n2;

15             n2=n1.next;

16         }

17         

18         return helper.next;

19     }

20 }

 第二遍：

 1 public class Solution {

 2     public ListNode swapPairs(ListNode head) {

 3         // Note: The Solution object is instantiated only once and is reused by each test case.

 4         if(head == null || head.next == null) return head;

 5         ListNode header = new ListNode(-1), first = head, second = head.next, cur = header;

 6         header. next = head;

 7         while(second != null ){

 8             first.next = second.next;

 9             second.next = first;

10             cur.next = second;

11             cur = first;

12             first = cur.next;

13             if(first != null)

14                 second = first.next;

15             else

16                 second = null;

17         }

18         return header.next;

19     }

20 }