【Leetcode_总结】19. 删除链表的倒数第N个节点- python

Q:

给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

示例：

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.

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链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/description/

思路：先求长度，然后再依次将节点相连，略过倒数第n个节点

代码：

# 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
        """
        stack=[]
        pre = head
        while pre.next:
            stack.append(pre)
            pre = pre.next
        stack.append(pre)
        if (len(stack)) == n:
            return head.next
        
        del_num = len(stack)-n
        head_ = new = stack.pop(0)
        new.next = None
        tmp = 1
        flag = True
        while stack:
            if tmp == del_num and flag:
                stack.pop(0)
                flag = False
            else:
                new.next = stack.pop(0)
                tmp += 1
                new = new.next
                new.next = None
        return head_