#19 Remove Nth Node From End of List ——Top 100 Liked Questions-优快云博客

本文链接：https://blog.youkuaiyun.com/suixuejie/article/details/90084681

本文介绍了一种高效算法，用于在一次遍历中移除链表的倒数第N个节点，并返回修改后的链表头。通过巧妙使用辅助变量，算法能够处理各种边界情况，包括当N等于链表长度时。该算法在Python实现中表现优秀，运行速度快于大部分在线提交。

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

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.

Follow up:

Could you do this in one pass?

"""

第一次：遍历链表中的每一个元素，用辅助变量t判断是否是倒数第n个，分三种情况：n<length,按部就班的找；n>length，返回整个链表；n=length，从链表的第二个开始返回。

"""
# 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
        """
        if head == None: return None
        if head.next == None and n == 1: return None
        prob = head
        while prob:
            p = prob
            t = n
            while t + 1:
                if t == 0 and p == None:                           #n=length
                    return head.next
                if t == 0 and p.next != None:                      #n<length
                    prob = prob.next
                if t == 0 and p.next == None:                      #n<length
                    prob.next = prob.next.next
                    return head
                if t!= 0 and p == None:                            #n>length
                    return head
                p = p.next
                t = t - 1
"""

Runtime: 24 ms, faster than 77.63% of Python online submissions for Remove Nth Node From End of List.

Memory Usage: 11.8 MB, less than 5.26% of Python online submissions for Remove Nth Node From End of List.

"""