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:
输入:head = [1,2,3,4,5], n = 2 输出:[1,2,3,5]
示例 2:
输入:head = [1], n = 1 输出:[]
示例 3:
输入:head = [1,2], n = 1 输出:[1]
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?
不是很懂为什么大佬会用一个假的头节点,防止快指针为NULL吗?
cpp
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* fast = head;
ListNode* slow = head;
while(n--) fast=fast->next;
if(fast == NULL) return head->next;
while(fast->next != NULL){
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;
return head;
}