【leetcode】Remove Nth Node From End of List

题目简述：

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：

标准的双指针，第一个指针先走n步，注意删除时候的处理

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
# @return a ListNode
def removeNthFromEnd(self, head, n):
posa = ListNode(0)
posa = head
posb = ListNode(0)
posb = head
pre = ListNode(0)
pre.next = head
while n > 0:
posa = posa.next
n -= 1
while posa != None:
#print 'a',posa.val
#print 'b',posb.val
posa = posa.next
posb = posb.next
pre = pre.next
#print posb.val
if pre.next == head:
head = head.next
else:
pre.next = pre.next.next

return head