Given a linked list, remove the n-th node from the end of list and return its head.
Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
if head is None or head.next is None:
return None
slow = fast = head
for i in range(n):
fast = fast.next
if fast is None:
head = head.next
return head
while fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head