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
/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; * 还是双指针的问题,关键的问题是边界的考虑。就是当n刚好是结点个数时的情况。 */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @param n int整型 * @return ListNode类 */ ListNode* removeNthFromEnd(ListNode* head, int n) { // write code here if (head == nullptr) return nullptr; auto slow = head; auto fast = head; for (int i = 0; i < n; ++i) fast = fast->next; if (fast == nullptr) { auto ret = head->next; delete head; return ret; } while (fast->next != nullptr) { slow = slow->next; fast = fast->next; } auto temp = slow->next; slow->next = temp->next; delete temp; return head; } };