目录
1.题目
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:[]
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
2.代码
/**
* 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* fast = head;
ListNode* slow = head;
ListNode* pre = new ListNode(-1);
pre->next = head;
ListNode* ans = pre;
int i = 0;
while(fast!=NULL)
{
if(i<n)
{
fast = fast->next;
i++;
}
else
{
fast = fast->next;
slow = slow->next;
pre = pre->next;
}
}
pre->next = slow->next;
return ans->next;
}
};