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;
}
};

