[LeetCode]Remove Nth Node From End of List@python

最新推荐文章于 2021-08-04 10:48:57 发布

benjamin_sunny_li

最新推荐文章于 2021-08-04 10:48:57 发布

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CC 4.0 BY-SA版权

文章标签：链表快慢指针

本文链接：https://blog.youkuaiyun.com/weixin_38111819/article/details/79115684

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。通过使用快慢指针技巧，确保算法仅需遍历一次链表即可完成操作。

Given a linked list, remove the n^th 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.

思路：设置快慢指针，快指针先跑，跑了n后再和慢指针一起跑。

特别注意n如果等于链表长度直接返回原链表，这里边有小心链表越界的情况，多码几遍代码注意一下！

# 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
        """
        fst = sec = head
        for _ in range(n):
            fst = fst.next
        if not fst:
            return head.next
        while fst.next:
            fst = fst.next
            sec = sec.next
        sec.next = sec.next.next
        return head