删除链表的倒数第N个结点
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:
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| 输入:head = [1], n = 1 输出:[]
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示例 3:
1 2
| 输入:head = [1,2], n = 1 输出:[1]
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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?
经典双指针。
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| class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode fast = head; ListNode slow = head; for(int i=0;i<n;i++){ fast=fast.next; } if(fast==null){ head = head.next; return head; } while(fast.next!=null){ fast=fast.next; slow=slow.next; } slow.next = slow.next.next; return head; } }
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2025/1/24
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| class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode p = head; int l = 0; while (p != null) { l++; p = p.next; } ListNode newHead = new ListNode(); newHead.next = head; p = newHead; for (int i = 0; i < l - n; i++) { p = p.next; } p.next = p.next.next; return newHead.next; } }
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