题目描述
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:[]
个人C++解答
GYL
本来想用辅助向量,写着写着发现没必要,不如直接用指针方便
/** * 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* result = new ListNode(); ListNode *p = head, *q = result; int len = 0, count = 0; while (p) { len++; p = p->next; } p = head; while (p) { count++; if (count == (len - n + 1)) break; q->next = p; p = p->next; q = q->next; } p = p->next; q->next = p; return result->next; } };