[leetcode: Python]19. Remove Nth Node From End of List

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.

解题思路：加一个头结点dummy，并使用双指针p1和p2。p1先向前移动n个节点，然后p1和p2同时移动，当p1.next==None时，此时p2.next指的就是需要删除的节点。

方法一：72ms

# 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
p1 = p2 = dummy
for i in range(n):
p1 = p1.next
while p1.next:
p1 = p1.next
p2 = p2.next
p2.next = p2.next.next

return dummy.next

方法二：55ms

# 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
"""
return self.helper(head, n)[1]

def helper(self, head, n):
if head == None:
return (0,None)
num, ref = self.helper(head.next, n)
head.next = ref

if num == n-1:
return (num + 1, ref)
else:
return (num + 1, head)

方法三：45ms

# 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
"""
p1=head
p2=head
for i in range(n):
p1=p1.next

if p1==None:
p2=p2.next
return p2

while p1.next!=None:
p1=p1.next
p2=p2.next

p2.next=p2.next.next

return head