leetcode 【 Remove Nth Node From End of List 】 python 实现

题目：

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.

代码：oj在线测试通过 Runtime: 188 ms

 # 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):

         if head is None:

             return head

         dummyhead = ListNode(0)

         dummyhead.next = head

         p1 = dummyhead

         p2 = dummyhead

         for i in range(0,n):

             p1 = p1.next

         while p1.next is not None:

             p1 = p1.next

             p2 = p2.next

             if p1.next is None:

                 break

         p2.next = p2.next.next

         return dummyhead.next

思路：

Linked List基本都需要一个虚表头，这道题主要思路是双指针

让第一个指针p1先走n步，然后再让p1和p2一起走；当p1走到链表最后一个元素的时候，p2就走到了倒数n+1个元素的位置；这时p2.next向表尾方向跳一个。

注意下判断条件是p1.next is not None，因此在while循环中添加判断p1.next是否为None的保护判断。

再有就是注意一下special case的情况，小白我的习惯是在最开始就把这种case都判断出来；可能牺牲了代码的简洁性，有大神路过也请拍砖指点。

巴特西

leetcode 【 Remove Nth Node From End of List 】 python 实现

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