Given a linked list, remove the nth node from the end of list and return its head.
For 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.
Try to do this in one pass.
Have you met this question in a real interview?
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this problem can be done by doing with a fast pointer and a slow pointer.
O(n) time
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
dummyHead = ListNode(None)
dummyHead.next = head
fast = slow = dummyHead
for _ in range(n):
if fast:
fast = fast.next
if n == 0 or not fast:
return dummyHead.next
while fast.next:
fast = fast.next
slow = slow.next
if slow.next:
slow.next = slow.next.next
return dummyHead.next