给你一个链表,删除链表的倒数第 n
个结点,并且返回链表的头结点。
示例 1:

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| 输入:head = [1,2,3,4,5], n = 2 输出:[1,2,3,5]
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示例 2:
示例 3:
提示:
- 链表中结点的数目为
sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
进阶:你能尝试使用一趟扫描实现吗?
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| class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode fast = dummy; ListNode slow = dummy; while(n-- != 0 && fast != null){ fast = fast.next; } if(fast == null) return head;
while(fast.next != null){ fast = fast.next; slow = slow.next; } slow.next = slow.next.next; return dummy.next;
} }
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