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.
Tips：给定一个链表，在不改变链表的值，只改变结点的指针的情况下完成相邻两个结点的交换。
如：Input：1->2->3->4
Output：2->1->4->3
思路：设置两个指针，first second，分别为当前指针的next以及next.next。
将 first.next指向second.next;（1->3）
然后将second.next指向first(2->1)。
再将当前指针后移：cur.next=second; cur = first;
package medium; import dataStructure.ListNode; public class L24SwapNodesInPairs { public ListNode swapPairs(ListNode head) { if (head == null) return head; ListNode node = new ListNode(0); node.next = head; ListNode cur = node; while (cur.next != null && cur.next.next!=null) { ListNode first = cur.next; ListNode second = cur.next.next; first.next = second.next;//1->3 second.next=first;//2->1->3 cur.next=second; cur = first; } return node.next; } public static void main(String[] args) { ListNode head1 = new ListNode(1); ListNode head2 = new ListNode(2); ListNode head3 = new ListNode(3); ListNode head4 = new ListNode(4); ListNode head5 = new ListNode(5); ListNode head6 = new ListNode(6); ListNode head7 = null; head1.next = head2; head2.next = head3; head3.next = head4; head4.next = head5; head5.next = head6; head6.next = head7; L24SwapNodesInPairs l24 = new L24SwapNodesInPairs(); ListNode head = l24.swapPairs(head1); while (head != null ) { System.out.println(head.val); head=head.next; } } }
