237. 删除链表中的节点

请编写一个函数，使其可以删除某个链表中给定的（非末尾的）节点，您将只被给予要求被删除的节点。

比如：假设该链表为 1 -> 2 -> 3 -> 4  ，给定您的为该链表中值为 3 的第三个节点，那么在调用了您的函数之后，该链表则应变成 1 -> 2 -> 4 。

代码如下：耗时60ms

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

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val = node.next.val
        node.next= node.next.next

19. 删除链表的倒数第N个节点

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.

代码如下： 耗时52ms

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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        ##### the first method is out of time
        
        List = []
        count = 0
        while (head):
            List.append(head)
            head = head.next
            count = count + 1

        if count == 1:
            return None
        if List[-n].next == None:
            List[-n - 1].next = None
            return List[0]
        else:
            List[-n].val = List[-n].next.val
            List[-n].next = List[-n].next.next
        return List[0]