This is my array and custom method to reverse an array output without using the reverse method. not sure where it broke, tried running it in console, no dice.
numbers = [1, 2, 3, 4, 5, 6]
def reversal(array)
do |item1, item2| item2 <=> item1
end
p reversal(numbers)
Here's one way to handle this. This is not very efficient but works.
def reversal(array)
reversed = []
loop do
reversed << array.pop
break if array.empty?
end
reversed
end
Here is another implementation that does the same thing:
def reversal(array)
array.each_with_index.map do |value, index|
array[array.count-index-1]
end
end
<, ==, and > into one operator; so, instead of returning true or false it returns -1 (for 'a is less than b), 0` (for 'a is equal to b') or +1 (for 'a is greater than b').
array: array = [1,2,3]; reversal(array) #=> [3,2,1]; array = []. I think you should either leave arr unchanged or replace it with the arr reversed (depending on West Coast Charlie's needs), but certainly not replace it with an empty array.So many ways... Here are three (#1 being my preference).
numbers6 = [1, 2, 3, 4, 5, 6]
numbers5 = [1, 2, 3, 4, 5]
For all methods my_rev below,
my_rev(numbers6)
#=> [6, 5, 4, 3, 2, 1]
my_rev(numbers5)
#=> [5, 4, 3, 2, 1]
#1
def my_rev(numbers)
numbers.reverse_each.to_a
end
#2
def my_rev(numbers)
numbers.each_index.map { |i| numbers[-1-i] }
end
#3
def my_rev(numbers)
(numbers.size/2).times.with_object(numbers.dup) do |i,a|
a[i], a[-1-i] = a[-1-i] , a[i]
end
end
there are so many ways to do this
a=[1,2,3,4,5,6,7,8]
i=1
while i <= a.length/2 do
temp = a[i-1]
a[i-1] = a[a.length-i]
a[a.length-i] = temp
i+=1
end
a=[1,2,3,4,5,6]
i=0
b=[]
t=a.length
while i< t do
b << a.pop
i+=1
end
a=[1,2,3,4,5,6]
b=[]
loop do
b << a.pop
break if a.empty?
end
a = [1,2,3,4,5]
b = []
a.length.times { |i| b << a[(i+1)*-1] }
b
=> [5,4,3,2,1]
We can implement the array reversal functionality without using built-in ruby methods.
arr = [1,2,3,4,5]
counter = 0
size = (arr.size/2.0).round(0)
while counter <= size - 1
first_el = arr[counter]
arr[counter] = arr[-(counter+1)]
arr[-(counter+1)] = first_el
counter += 1
end
p arr
This sample code will shift the array elements without creating a new array.
<=>does?[1,3,2], is the correct output[3,2,1]or[2,3,1]?doand thedo |stuff|construct are part of a method call, not a method definition. Of course, you can call a method from within the body of another method, but you're not doing that. Also, everydoneeds a correspondingendregardless.[1,2,3,4,5,6]is already in order, so putting it backwards from what it is now and backwards from 'in order' is the same thing. What if you had a list that was[1,2,4,5,6,3]? Would you want the result to be[6,5,4,3,2,1]or[3,6,5,4,2,1]?defand ends withend, whereas a method invocation is where you call the method. Fromdef reversal(array)toendis a method definition (though not a valid one). The last line is a method call (actually two method calls, one call toreversaland one call top). Method calls can include a 'block' (though many don't), which is a bunch of code betweendoandendor between{and}. Like in @roychri's answer below, he passes a block to the methodloop