一.题目要求
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。
二.题目难度
中等
三.输入样例
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:[]
示例 3:
输入:head = [1,2], n = 1
输出:[1]
提示:
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
进阶:你能尝试使用一趟扫描实现吗?
四.解题思路
解法1:两次扫描
解法2:快慢指针
五.代码实现
解1
/**
* 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* myhead = new ListNode(-1);
myhead->next = head;
ListNode* p = myhead;
int len = 0;
while(head)
{
len++;
head = head->next;
}
int nownode = 0;
while(nownode != len - n)
{
p = p->next;
nownode++;
}
p->next = p->next->next;
return myhead->next;
}
};
解2
/**
* 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* myhead = new ListNode(-1);
myhead->next = head;
ListNode* slow = myhead;
ListNode* quick = slow;
while(n>0)
{
quick = quick->next;
n--;
}
while(quick->next)
{
quick = quick->next;
slow = slow->next;
}
slow->next = slow->next->next;
return myhead->next;
}
};
六.题目总结
链表的问题多想想快慢指针