Question

在尝试构建完全递归的 LinkedList 反转时，如何避免嵌套输出 [(((4, 3), 2), 1)]？我想将 [4, 3, 2, 1)] 作为输出，并想知道这在 Python 中是否可行，而无需迭代最终结果。

class Leaf:
    def __init__(self, value):
        self.value = value
        self.next_node = None
    def set_next(self, next_node):
        self.next_node = next_node

class LinkedList:
    def __init__(self, list_init):
        self.len = len(list_init)
        self.node = Leaf(list_init[0])
        self.node.set_next(build_node(list_init[1:]))
        
    def print(self):
        print("{}".format(self.node.value))
        next_node = self.node.next_node
        while(next_node is not None):
            print("{}".format(next_node.value))
            next_node = next_node.next_node
    
def build_node(list_nodes):
    if len(list_nodes) > 1:
        node = Leaf(list_nodes[0])
        node.set_next(build_node(list_nodes[1:]))
    else:
        node = Leaf(list_nodes[0])    
    return node

linkedList = LinkedList(list_init = [1, 2, 3, 4])

def reverse_linked_list(linkedList):
    first_node = linkedList.node
    def _reverse_linked_list(node):
        print("Enter: {}".format(node.value))
        if node.next_node is not None:
            print("Call: {}".format(node.next_node.value))
            return _reverse_linked_list(node.next_node), node.value
        return node.value
    return [_reverse_linked_list(first_node)]

reverse_linked_list(linkedList)

Answer 1

我认为您可以将函数更新为：

def reverse_linked_list(linkedList):
    first_node = linkedList.node

    def _reverse_linked_list(node):
        print("Enter: {}".format(node.value))
        if node.next_node is not None:
            print("Call: {}".format(node.next_node.value))
            prev_value = _reverse_linked_list(node.next_node)
            if isinstance(prev_value,int):
                prev_value = (prev_value,)
            return (*prev_value, node.value)
        return node.value

    return [*_reverse_linked_list(first_node)]

reverse_linked_list(linkedList)

反转 LinkedList 的嵌套输出

1 个答案: