给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
示例:
给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后,链表变为 1->2->3->5.
说明:
给定的 n 保证是有效的。
进阶:
你能尝试使用一趟扫描实现吗?
public class RemoveNthFromEnd {
@Test
public void removeNthFromEndTest() {
Assert.assertNull(removeNthFromEnd(new ListNode(1), 1));
ListNode listNode = new ListNode(1);
listNode.next = new ListNode(2);
Assert.assertEquals(2, removeNthFromEnd(listNode, 2).val);
}
@Test
public void removeNthFromEndTest1() {
ListNode listNode = new ListNode(1);
listNode.next = new ListNode(2);
Assert.assertEquals(1, removeNthFromEnd(listNode, 1).val);
}
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode temp = head;
Stack<ListNode> stack = new Stack<>();
while (head != null) {
stack.push(head);
head = head.next;
}
for (int i = 0; i < n-1; i++) {
stack.pop();
}
ListNode targetNode = stack.pop();
if (stack.size() == 0) {
temp = targetNode.next;
} else if(n == 1) {
ListNode lastNode = stack.pop();
lastNode.next = null;
} else {
targetNode.val = targetNode.next.val;
targetNode.next = targetNode.next.next;
}
return temp;
}
}