Given a linked list, remove the n-th node from the end of list and return its head.
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.
Given n will always be valid.
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 """ nth_prev = head tail = head for i in range(0, n): tail = tail.next if tail is None: head = head.next return head while tail.next is not None: nth_prev = nth_prev.next tail = tail.next nth_prev.next = nth_prev.next.next return head