题目描述
给定一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。
Example1:
输入：head = [1,2,3,4,5], n = 2 输出：[1,2,3,5]
Example2:
输入：head = [1], n = 1 输出：[]
Example3:
输入：head = [1,2], n = 1 输出：[1]
Note:
- 链表中结点的数目为 sz
- 1 <= sz <= 30
- 0 <= Node.val <= 100
- 1 <= n <= sz
解题思路
/**
* 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) {
if(head == nullptr){
return nullptr;
}
ListNode* fast = head, *slow = head;
while(fast && n){
n--;
fast = fast -> next;
}
if(fast == nullptr){
return head -> next;
}
while(fast -> next != nullptr){
fast = fast -> next;
slow = slow -> next;
}
slow -> next = slow -> next -> next;
return head;
}
};
