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May 4, 2017 02:38
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Linked list reversal solutions
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# A node of a linked list | |
class Node | |
attr_reader :value, :next_node | |
def initialize(value, next_node = nil) | |
@value = value | |
@next_node = next_node | |
end | |
def reverse(reversed = nil) | |
node = Node.new value, reversed | |
next_node ? next_node.reverse(node) : node | |
end | |
def reverse!(node = self, reversed = reverse) | |
@value = reversed.value | |
next_node ? next_node.reverse!(node, reversed.next_node) : node | |
end | |
def to_s | |
"<#{object_id} #{@value}> -> #{next_node.nil? ? "nil\n" : next_node.to_s}" | |
end | |
end | |
linked_list = Node.new(1, Node.new(2, Node.new(3, Node.new(4)))) | |
puts "LinkedList: #{linked_list}" | |
puts "Non-mutate Reverse: #{linked_list.reverse}" | |
puts "Mutate Reverse: #{linked_list.reverse!}" |
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Output:
Note: digits in between
<>
represent theobject_id
andvalue
i.e. proving that the object is the same or different object in memory .