leetcode-easy-listnode-19 remove nth node from end of list

mycode  88.29%

关键是一定要head前新建一个节点，否则要分类讨论很多次来避免slow或者fast出现None.next的错误

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

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        dummy = ListNode(-1)
        dummy.next = head
        slow = fast = dummy
        #print(n,head.val)
        while n:
            fast = fast.next
            n -= 1
        #print(n,head.val,fast)
        while fast and fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy.next

例如，下面这种情况就要分情况讨论，但是会更快

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

class Solution:
    def removeNthFromEnd(self, head, n):
        fast = slow = head
        for _ in range(n):
            fast = fast.next
        if not fast:   #例如1-》2-》3-》4-》None，n=4或者5的时候，删除的就应该是第一个节点，所以返回head.next就好
            return head.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return head