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 }