leetcode19. 删除链表的倒数第 N 个结点

思路
获取链表长度，然后开始遍历，当遍历到 链表长度-n-1时，此时结点的下一个结点即为要删除的结点，所以停止循环，直接删下一个结点 即p.next=p.next.next
需要注意一个特殊情况：删除头结点

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if head is None:
            return None
        if head.next is None:
            return None
        p = head
        length = 0
        while p:
            length+=1
            p=p.next
        length=length-n
        p =head
        i = 0
        while p:
            i+=1
            if i<length:
                p=p.next
            else:
                break
        if length== 0:
            #删除头结点
            p.val =p.next.val
            p.next=p.next.next
        else:
            p.next =p.next.next
        return head

若不需要特别注意删除头结点：

创建一个新链表=[0] +原链表 此时进行上方操作

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if head is None:
            return None
        if head.next is None:
            return None
        p = head
        length = 0
        while p:
            length+=1
            p=p.next
        length=length-n
        prelist =ListNode(0)
        pre = prelist
        pre.next=head
        p = prelist
        i = 0
        while p:
            i+=1
            if i<=length:
                p=p.next
            else:
                break
        p.next=p.next.next
        return prelist.next