head of a linked list, remove the
nth node from the end of the list and return its head.
Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5]
Input: head = , n = 1 Output: 
Input: head = [1,2], n = 1 Output: 
- The number of nodes in the list is
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
Follow up: Could you do this in one pass?
# 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: p1 = head p2 = head for _ in range(n): if p2: p2 = p2.next else: break if p2==None: return head.next while p2.next: p1 = p1.next p2 = p2.next p1.next = p1.next.next return head