Given the head of a linked list, remove the nth node from the end of the list and return its head.
Example 1:
Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5]
Example 2:
Input: head = [1], n = 1 Output: []
Example 3:
Input: head = [1,2], n = 1 Output: [1]
Constraints:
sz.1 <= sz <= 300 <= Node.val <= 1001 <= n <= sz
Follow up: Could you do this in one pass?
/**
* 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) {
auto lt=head, rt=head;
while(rt && n--)
rt=rt->next;
if(!rt){
if(!n)
return head->next;
return head;
}
while(rt->next){
lt=lt->next;
rt=rt->next;
}
lt->next=lt->next->next;
return head;
}
};/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0, head), fast = dummy, slow = dummy;
while (n-->=0)
fast = fast.next;
while (fast != null){
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return dummy.next;
}
}