题解 | #删除链表的倒数第n个节点#

指定头结点的前一个节点为None，赋值给pre，当前头结点赋值给fast指针
利用快慢指针找到删除节点的前一个节点

class Solution:
    def removeNthFromEnd(self , head , n ):
        # write code here
        fast = head
        newHead = ListNode(None)
        newHead.next = head
        pre = newHead
        for i in range(n):
            fast = fast.next
        while fast is not None:
            fast = fast.next
            pre = pre.next
        # del pre next
        pre.next = pre.next.next
        return newHead.next