【LeetCode with Python】 Remove Nth Node From End of List

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原题页面： https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
题目类型：
难度评价：★
本文地址： http://blog.youkuaiyun.com/nerv3x3/article/details/36897851

Given a linked list, remove the n^th 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.

常见的快慢指针题目。快指针领先慢指针n步，这样当快指针到达尾部时，慢指针正好指向需要删除的节点的父结点。
注意由于要删除的结点可能刚好是head节点，这种情况与其特殊处理，不如加一个“伪head”，这也是类似的链表类题目的通用做法。

class Solution:
    # @return a ListNode
    def removeNthFromEnd(self, head, n):
        if None == head:
            return head
        if None == head.next:
            return None
        new_head = ListNode(-1)
        new_head.next = head
        fast = new_head
        for i in range(0, n):
            if None != fast.next:
                fast = fast.next
            else:
                return head
        slow = new_head
        while None != fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return new_head.next