python - 将 n 个值插入一个空的双向链表(伪代码)

Question

有人可以帮我验证我的伪代码是否正常工作，以便提供的函数将 n 个数字附加到列表中吗？

我在第一年的数据科学作业中需要帮助。

我们正在学习数据结构和双向链表。我理解我认为的双向链表背后的逻辑，但是我很难用伪代码编写发生的事情。

"Assignment 1: Let the list S be empty as a start. Now input n (natural numbers, e.g.: 1, 2, 3 ...) one at the time. When the last number of n is placed into the list, the list will be a sorted list containing the n numbers. Describe why we'll get a sorted n number in O(n^2) time."

我的作业答案写在下面，我真的不确定它是否正确。

// Our nodes will consist of 3 cells in each object.
// key  = a number (int)
// prev = address pointer to previous node 
// next = address pointer til next node

// This function creates an empty list
function emptyList()
   L = new List{head = nil, size = 0}
   return L

// This function creates a node.
function makeNode(val)
  node = new Node{prev = NIL, key = val, next = NIL}
  return node

// This function inserts n amount of nodes to an empty list
function InsertNodes(n)

    // Create an empty list, S.
    emptyList()

    // Initiate the first node
    S.head  = makeNode(1) 

    for i = 2 to n-1

        prevNode = makeNode(i-1)
        newNode  = makeNode(i)

        while newNode.prev == NIL do

            // connect addresses of nodes
            prevNode.next = newNode.prev
            newNode.prev  = prevNode.next

类(class)是关于算法和离散数学的

Answer 1

// Our nodes will consist of 3 cells in each object.
// key  = a number (int)
// prev = address pointer to previous node 
// next = address pointer til next node

// This function creates an empty list
function emptyList()
   L = new List{head = nil, size = 0}
   return L

// This function creates a node.
function makeNode(val)
  node = new Node{prev = NIL, key = val, next = NIL}
  return node

// This function inserts n amount of nodes to an empty list
function InsertNodes(n)

    // Create an empty list, S.
    emptyList()

    // Initiate the first node
    S.head  = makeNode(1) 

    //tail keeps the last node
    tail = head
    for i = 2 to n

        tail.next = makeNode(i)
        tail.next.prev = tail
        tail = tail.next