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:
- The number of nodes in the list is
sz. -
1 <= sz <= 30 -
0 <= Node.val <= 100 -
1 <= n <= sz
Follow up: Could you do this in one pass?
SOLUTION:
# 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, n):
slow = head
fast = head
for _ in range(n):
fast = fast.next
while fast and fast.next:
slow = slow.next
fast = fast.next
if slow == head and not fast:
return head.next
slow.next = slow.next.next
return head
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