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Given the head
of a linked list, remove the nth
node from the end of the list and return its head.
Follow up: Could you do this in one pass?
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 <= 30
0 <= Node.val <= 100
1 <= n <= sz
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
dummy = ListNode(next=head)
p = q = dummy
for i in range(n):
p = p.next
while p.next:
p, q = p.next, q.next
q.next = q.next.next
return dummy.next
/**
* 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);
ListNode p = dummy, q = dummy;
while (n-- > 0) {
p = p.next;
}
while (p.next != null) {
p = p.next;
q = q.next;
}
q.next = q.next.next;
return dummy.next;
}
}
/**
* 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 = new ListNode(0, head);
ListNode* p = dummy;
ListNode* q = dummy;
while (n-- > 0) {
p = p->next;
}
while (p->next != nullptr) {
p = p->next;
q = q->next;
}
q->next = q->next->next;
return dummy->next;
}
};
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd(head *ListNode, n int) *ListNode {
dummy := &ListNode{Val:0, Next:head}
p := dummy
q := dummy
for n > 0 {
p = p.Next
n--
}
for p.Next != nil {
p = p.Next
q = q.Next
}
q.Next = q.Next.Next
return dummy.Next
}
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