【LEETCODE】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.

参考： http://yucoding.blogspot.com/2013/04/leetcode-question-82-remove-nth-node.html
题意：
给一个linked list和数字n，移除倒数第n个数，返回head

思路：
两个指针p，q同时从head出发
q先移动n步
然后p，q同时移动
当q到达end时，p.next=p.next.next即可

# 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
"""

head1=ListNode(0)             #增加0点
head1.next=head
head=head1

p=q=head

for i in range(n+1):          #因为多了一个0点，所以q要走n＋1步
q=q.next

while q:                      #因为多了一个0点，所以是判断q而不是q.next，这样p才可以也多走一步
p=p.next
q=q.next

p.next=p.next.next

return head.next