19. Remove Nth Node From End of List

博客围绕LeetCode题目展开,关键在于实现一次遍历操作。若不要求一次遍历,可先求链表长度再定位节点。以示例说明,通过移动节点位置确定要删除的节点。同时考虑特殊情况,如head为None时的处理,并给出了AC代码。

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思路

本题关键在于do this in one pass,否则可以直接第一遍遍历求List的长度,再求n-th node from the start of list。

首先假设List的长度是N,则移动N次后ListNode为none,以题中示例为例,n=2,N=5时,ListNode移动两次之后,还有三次ListNode为none,而我们需要去掉的数恰好就是从头开始移动三位之后的数。

为缩短一次移动我们用head.next是否为空进行判断:

        while(head.next):
            head = head.next
            head1 = head1.next

但又会出现问题,比如这一点输入:

[1,2,3,4,5]
5

当head是None时,head.next会报错,对这种情况我们单独处理

        if head == None:
            return head1.next

AC代码

# 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
        """
        head1 = head
        head2 = head
        for _ in range(n):
            head = head.next
        if head == None:
            return head1.next
        while(head.next):
            head = head.next
            head1 = head1.next
        head1.next = head1.next.next
        return head2
        

 

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