Lintcode 174. Remove Nth Node From End of List (Easy) (Python)

2Young2Simple

于 2018-07-31 00:13:43 发布

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文章标签： Lintcode Python

本文链接：https://blog.youkuaiyun.com/weixin_41677877/article/details/81292076

本文介绍了一种高效算法，用于从链表中移除倒数第N个节点。通过双指针技巧，在不获取链表长度的情况下完成操作。提供了完整的Python代码实现。

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Remove Nth Node From End of List

Description:

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

Example

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

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

Challenge

Can you do it without getting the length of the linked list?

Code:

"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""

class Solution:
    """
    @param head: The first node of linked list.
    @param n: An integer
    @return: The head of linked list.
    """
    def removeNthFromEnd(self, head, n):
        # write your code here
        if head.next == None and n == 1:
            return None
        ad = ListNode(0)
        ad.next = head
        ind0 = ad
        ind1 = ad
        for i in range(n):
            ind1 = ind1.next
        while ind1.next != None:
            ind1 = ind1.next
            ind0 = ind0.next
        tmp = ind0.next
        ind0.next = tmp.next
        tmp.next = None
        return ad.next