[leetcode]Remove Nth Node From End of List @ Python

原题地址：http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

题意：

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.

这道题的含义是删除链表的倒数第n个节点。

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。

代码：

# 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):

        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

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