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.
# 不带头
# 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
"""
if head == None:
return None
list_node = []
temp_node = head
while temp_node!=None:
list_node.append(temp_node)
temp_node = temp_node.next
l_length = len(list_node)
curr_node = list_node[l_length-n]
if l_length-n > 0:
prev_node = list_node[l_length-n-1]
prev_node.next = curr_node.next
else:
head = curr_node.next
return head# 带头
# 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
"""
if head == None:
return None
list_node = []
fake_head = ListNode(0)
fake_head.next = head
temp_node = fake_head
while temp_node!=None:
list_node.append(temp_node)
temp_node = temp_node.next
l_length = len(list_node)
curr_node = list_node[l_length-n]
prev_node = list_node[l_length-n-1]
prev_node.next = curr_node.next
return fake_head.next
# 带头
# 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
"""
if head == None:
return None
fake_head = ListNode(0)
fake_head.next = head
p1 = p2 = fake_head
for i in range(n):
if p2==None: return None
p2 = p2.next
while p2.next!=None:
p2 = p2.next
p1 = p1.next
p1.next = p1.next.next
return fake_head.next