题目说明

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

输入：head = [1,2,3,4,5], n = 2

输出：[1,2,3,5]

解题思路

1. 遍历链表A到尾部，并计算count；然后遍历链表B判断count>n，跳出循环则head.next=head.next.next。
2. 遍历链表A count与n的距离，然后接着遍历链表A，链表B，跳出循环则head.next = head.nexe.next

代码如下

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:

    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        # 通过链表移动计算出链表大小与删除节点位置的距离
        res = ListNode(next=head)  # 创建一个虚拟节点,next指向头
        h_head = l_head = res
        count = 0
        while l_head.next is not None and count != n:  # 移动尾节点，使其与头节点保持n个节点
            l_head = l_head.next
            count += 1
        while l_head.next is not None:  # 尾节点遍历完
            l_head = l_head.next
            h_head = h_head.next
        h_head.next = h_head.next.next  # 头节点的下一个节点指向它的下下个节点
        return res.next

    def removeNthFromEnd_1(self, head: ListNode, n: int) -> ListNode:
        # 先遍历链表长度，然后判断长度与删除节点的距离
        res = ListNode(next=head)  # 创建一个虚拟节点,next指向头
        h_head = l_head = res
        count = 0
        while l_head.next is not None:  # 第一位是虚拟的头节点
            l_head = l_head.next
            count += 1
        while count > n:
            h_head = h_head.next
            count -= 1
        h_head.next = h_head.next.next  # 头节点的下一个节点指向它的下下个节点
        return res.next

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总结