如何在 Ruby 中反转链表 [英] How to reverse a linked list in Ruby
问题描述
在下面的突变示例中,我不明白链表是如何反转的.
In the mutation example below, I don't understand how the linked list is reversed.
class LinkedListNode
attr_accessor :value, :next_node
def initialize(value, next_node=nil)
@value = value
@next_node = next_node
end
end
def print_values(list_node)
print "#{list_node.value} --> "
if list_node.next_node.nil?
print "nil
"
return
else
print_values(list_node.next_node)
end
end
def reverse_list(list, previous=nil)
current_head = list.next_node
list.next_node = previous
if current_head
reverse_list(current_head, list)
else
list
end
end
node1 = LinkedListNode.new(37)
node2 = LinkedListNode.new(99, node1)
node3 = LinkedListNode.new(12, node2)
print_values(node3)
puts "-------"
revlist = reverse_list(node3)
print_values(revlist)
如果我只返回 current_head
,我会得到 99->37->nil
,这是有道理的,因为 99
将是 <代码>next_node.返回下一行,
If I just return current_head
, I get 99->37->nil
, which makes sense because 99
would be next_node
. Returning the next line,
list.next_node = previous
抛出错误,因为print_values
方法无法打印nil
的值.我不明白是什么颠倒了列表.如果有人能向我解释这一点,我将不胜感激.
throws an error because print_values
method can't print the value for nil
. I'm not understanding what is reversing the list. If anyone could explain this to me, I would appreciate it.
推荐答案
这是我制作的一个小可视化.
Here's a little visualization I made up.
^
指向列表的头部.在递归的每一级,它的右箭头转向"以从右边的元素指向左边的元素.继续直到有一个右箭头(指向一个非零).如果右箭头指向 nil,则返回当前头部.
^
points to head of the list. At each level of recursion, its right arrow is "turned" to point from element on the right to element on the left. Proceed until there is a right arrow (pointing to a non-nil). If right arrow points to nil, return the current head.
previous
↓
nil 12 -> 99 -> 37 -> nil
^
previous
↓
nil <- 12 99 -> 37 -> nil
^
previous
↓
nil <- 12 <- 99 37 -> nil
^
nil <- 12 <- 99 <- 37
^
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