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.
use a stack to record nodes. and pop out n+1 times to get the n+1th node from end. Then, delete the nth node
# 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
"""
dummy = ListNode(0)
dummy.next = head
stack = []
l = -1
current = dummy
while current:
stack.append(current)
current = current.next
l += 1
node = 0
for _ in xrange(n+1):
node = stack.pop()
node.next = node.next.next
return dummy.next
