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:
sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
/**
* 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* dummy_head = new ListNode();
dummy_head->next = head;
ListNode* cur = dummy_head;
int _size = 0;
while (cur->next != NULL)cur = cur->next,_size++;
n = _size - n;
cur = dummy_head;
while (n--){
cur = cur->next;
}
cur->next = cur->next->next;
return dummy_head->next;
}
};