LeetCode----Remove Nth Node From End of List

最新推荐文章于 2021-01-01 13:13:30 发布

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最新推荐文章于 2021-01-01 13:13:30 发布

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分类专栏： Python LeetCode 文章标签： python LeetCode

本文链接：https://blog.youkuaiyun.com/whiterbear/article/details/50763164

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点。通过构建一个长度为N的尺子进行一次遍历即可完成操作，详细解析了尺子的构建与使用过程。

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Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

分析：

构建一个尺子，尺子长度为n，尺子顺着链表遍历，当尺子尾到达链表尾的时候，尺子头部记录要删除的结点，One pass算法。

代码：

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        # 构建一个长度为n的尺子
        rulertail = head
        rulerhead = head
        rulerheadpre = head
        flag = True
        for i in range(n - 1):
            rulertail = rulertail.next
        while rulertail.next is not None:
            rulertail = rulertail.next
            rulerhead = rulerhead.next
            if flag:
                flag = False
            else:
                rulerheadpre = rulerheadpre.next
        # 删除倒数第n个元素
        if rulerhead == rulertail:
            if head == rulerhead:
                return None
            rulerheadpre.next = None
        else:
            rulerhead.val = rulerhead.next.val
            rulerhead.next = rulerhead.next.next
        return head