如何在Python3中从头开始在本地实现和运行Leetcode问题（LinkedList、Trees）？

Question

我想在本地机器上实现 Leetcode 问题，因为我是初学者，并且想掌握 python 编程。我认为在本地解决问题有助于我详细地理解问题。这就是 Leetcode 206 的起点。反转链表.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
    ## your code goes here...

        

leetcode 平台上有很多 YT 视频解释该算法以及如何解决该算法，但没有一个视频教如何从头开始实现它。 所以，我尝试做以下事情：

##
from typing import List

class ListNode:
    def __init__(self, head = None, next = None):
        self.head = head
        self.next = next
    
    def __repr__(self):
        return self.head

## 
class Solution:
    def __init__(self):
        self.head = ListNode()
        
    def reverseList(self, nums: ListNode)-> ListNode:
        prev, cur = None, self.head
        
        while cur:
            temp = cur.next
            cur.next = prev
            prev = cur
            cur = temp
        return prev

当我在leetcode中提交时，解决方案本身是正确的。但是当我按照以下方式实现自己的 main 时，我得到以下结果。

## main1
if __name__ =="__main__":
    nums = [1,2,3,4,5]
    sol = Solution()
    ans = sol.reverseList(nums)
    print(ans)
            

输出如下所示：

<__main__.ListNode object at 0x000002E6E5CFB940>

您能帮助我如何在本地正确实现 main 并查看对象数据类型吗？

我面临树和列表节点类型问题的问题。 非常感谢。

Answer 1

from typing import Optional
class ListNode:
    def __init__(self, data=None):
        self.data = data
        self.next = None


# This is for converting a list to a linked list. 
# The output is the head of the linked list
def list_to_linked_list(values):
    head = None
    tail = None
    for val in values:
        node = ListNode(val)
        if head is None:
            head = tail = node
        else:
            tail.next = node
            tail = node
    return head


# This is for converting a linked list to a list. The input is the head. 
# The output is the list converted.
def linked_list_to_list(head):
    curr = head
    res = []
    while curr:
        res.append(curr.data)
        curr = curr.next
    return res


class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        prev, curr = None, head
        while curr:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        return prev


head = list_to_linked_list([0, 1, 2, 3])
solution = Solution()
print(linked_list_to_list(solution.reverseList(head)))
# Output: [5, 4, 3, 2, 1]