19. Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

Biglethz

最新推荐文章于 2024-01-12 15:01:11 发布

阅读量174

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/Biglethz/article/details/79706093

LeetCode 专栏收录该内容

10 篇文章

订阅专栏

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。该方法仅需遍历一次链表，通过使用两个指针实现。详细解释了特殊情况处理及算法步骤。

题目

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.

思路

两个指针first 、second 指向链表首部，首先移动first 指针n次，然后同时移动两个指针直到first 到达链表尾部

需要处理n等于链表长度的特殊情况

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
def removeNthFromEnd(head, n):
    """
    :type head: ListNode
    :type n: int
    :rtype: ListNode
    """
    if not head.next: return head.next
    first = head
    second = head
    while(first.next is not None):
        first = first.next
        if n > 0:
            n -= 1
            print(n,':',first)
            #处理n等于链表长度的特殊情况
            if n==1 and first.next is None:
                print('yes')
                return head.next
        else:
            second = second.next
    second.next =  second.next.next
    return head