反向第二半链表
例如:
偶数: 2-> 1-> 3-> 4-> 5-> 6-> 7-> 8 == ===> 2-> 1-> 3-> 4-> 8-> 7-> 6-> 5;反向链接列表的右半奇数:5-> 7-> 8-> 6-> 3-> 4 - > 2 ==> 5-> 7-> 8-> 2-> 4-> 3->如图6所示,中间 一个也需要被反转
class ListNode
{
int val;
ListNode next;
ListNode(int x) { val = x; }
}
class ReverseRightHalfLinkedList
{
public static void main(String[] args)
{
ListNode node1 = new ListNode(1);
ListNode node2 = new ListNode(2);
ListNode node3 = new ListNode(3);
ListNode node4 = new ListNode(4);
ListNode node5 = new ListNode(5);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;
ListNode res = reverse(node1);//line 31
// ListNode node = node1;
// while (node != null)
// {
// System.out.println(node.val);
// node = node.next;
// }
}
public static ListNode reverse(ListNode start)
{
int counter = 0;
ListNode node = start;
ListNode pre = start;
while (node!= null)
{
counter += 1;
node = node.next;
}
for (int i=0; i<counter/2; i++)
{
pre = start;
start = start.next;
}
ListNode cur = start;
if (counter%2 ==0)
{
while (cur != null)
{
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
}
else
{
pre = pre.next;
cur = start.next;
System.out.println(pre.val);
System.out.println(cur.val);
while (cur != null)
{
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
System.out.println("-----");
System.out.println(pre.val); // line 90
System.out.println(cur.val);
System.out.println("-----");
System.out.println();
}
}
return start;
}
}
首先,我收到错误消息。
在 ReverseRightHalfLinkedList.main(OA2.java:31)在 ReverseRightHalfLinkedList.reverse(OA2.java:90)线程 “主” 显示java.lang.NullPointerException异常
其次,我试图打印反向链表的顺序,它仍然是有序的。它没有被扭转。
请帮我弄清楚这两个问题。非常感谢!
非常感谢你的帮助。根据你的想法,我得到了一个更简洁的代码。再一次,非常感谢你的帮助。 – OregonDuck