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) {} * }; */ 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; } if (n <= 0 || n > 1000) { return nullptr; } ListNode* preHead = new ListNode(-1); preHead->next = head; ListNode* first = head; ListNode* last = head; ListNode* pre = preHead; for(int i = 1; i < n; i++) { if(last->next == nullptr) { return nullptr; } last = last->next; } // last为第n个节点 while (last->next != nullptr) { pre = pre->next; first = first->next; last = last->next; } // first为倒数第n个节点 // 删除first pre->next = first->next; first->next = nullptr; delete(first); return preHead->next; } };