删除链表的倒数第N个节点
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:
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?
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C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
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 == nullptr){
return head->next;
}
while(fast->next != nullptr){
slow = slow->next;
fast = fast->next;
}
slow->next = slow->next->next;
return head;
}
};
key:关键在于,走了n次,剩下length-n次,我要让fast停在链表结尾处,这样slow在倒数第n+1个位置,以防处理不了空指针问题