边工作边刷题：70天一遍leetcode: day 38-2

Remove Nth Node From End of List

要点

• first和second指针的距离：n是从1开始计数的，所以second指向的node和被删的node之间共n个node（包括这两个边界node）。first指向被删node的前一个，这样才能进行list node的删除
• 根据上面的距离关系，first开始指向dummy，second要首先从head开始单独移动n-1步，之后移动second到最后一个node，这样first.next指向被删node。所以用second.next!=null作为循环条件

# 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(0)
        dummy.next = head
        first, second = dummy, head
        while second.next and n>1:
            second=second.next
            n-=1
        
        if n!=1:
            return None
        
        while second.next:
            first=first.next
            second=second.next
            
        first.next=first.next.next
        return dummy.next