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:
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- 输入:head = [1,2,3,4,5], n = 2
- 输出:[1,2,3,5]
示例 2:
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- 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
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- /**
- * 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;
- }
- };