题解 | #删除链表的倒数第n个节点#
删除链表的倒数第n个节点
https://www.nowcoder.com/practice/f95dcdafbde44b22a6d741baf71653f6
使用虚拟头节点去删除
slow->next=slow->next->next;
fast到达第n个节点还需要再到下一步,使得slow指向需要删除节点的上一个节点
/** * struct ListNode { * int val; * struct ListNode *next; * }; */ class Solution { public: /** * * @param head ListNode类 * @param n int整型 * @return ListNode类 */ ListNode* removeNthFromEnd(ListNode* head, int n) { // write code here if(head==nullptr || n==0){ return head; } ListNode* vh=new ListNode(0); vh->next=head; ListNode* fast=vh , *slow=vh; for(int i=0; i< n; i++){ fast=fast->next; if(fast==nullptr){//如果越界 return head; } } fast=fast->next; while(fast!=nullptr){ fast=fast->next; slow=slow->next; } slow->next=slow->next->next; return vh->next; } };