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Programming-Idioms

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  • Kotlin

Idiom #19 Reverse a list

Reverse the order of the elements of the list x.
This may reverse "in-place" and destroy the original ordering.

x = x.reversed()

Immutable lists but not in-place.
val reversedView = x.asReversed()

Returns a reversed read-only view of the original List. All changes made in the original list will be reflected in the reversed one.
x.reverse()

MutableLists or Arrays can be reversed in place. 
with Ada.Containers.Vectors;
use Ada.Containers;
X.Reverse_Elements;
New implementation...
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programming-idioms.org 
x = x.reversed()
val reversedView = x.asReversed()
x.reverse()
X.Reverse_Elements;
int *p1 = x;
int *p2 = x + N-1;

while (p1 < p2)
{
    int temp = *p1;
    *(p1++) = *p2;
    *(p2--) = temp;
}
List.rev x
(reverse x)
std::reverse(begin(x), end(x));
x.Reverse();
reverse(x);
auto y = x.retro;
x = x.reversed.toList();
x = Enum.reverse(x)
lists:reverse(List)
  a = a(ubound(a,dim=1)::-1)
for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
	x[i], x[j] = x[j], x[i]
}
slices.Reverse(x)
func reverse[T any](x []T) {
	for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
		x[i], x[j] = x[j], x[i]
	}
}
x = x.reverse()
reverse x

x = x.reverse();
Collections.reverse(x);
reverse(x);
int i, m = x.size(), n = m-- / 2;
for (i = 0; i < n; ++i) swap(x, i, m - i);
static <T> void reverse(List<T> x){
	int n = x.size();
	for(int i=0;i<n/2;i++){
		T tmp = x.get(i);
		x.set(i, x.get(n-i-1));
		x.set(n-i-1, tmp);
	}
}
Collections.reverse(x);
(reverse x)
function array_reverse(x)
  local n, m = #x, #x/2
  for i=1, m do
    x[i], x[n-i+1] = x[n-i+1], x[i]
  end
  return x
end

rev = {}
for i=#x, 1, -1 do
	rev[#rev+1] = x[i]
end

-- in-situ reversal
function reverse(t)
  local n = #t
  local i = 1
  for i = 1, n do
    t[i],t[n] = t[n],t[i]

    n = n - 1
  end
end
rev = {}
for i=#x, 1, -1 do
	rev[#rev+1] = x[i]
end
x = rev
x.reverseObjectEnumerator.allObjects
$x = array_reverse($x, true);
var n,i: integer;
[...]
n:= length(x);
setlength(x, n+1);
for i := 0 to (n div 2)-1 do
	begin
		x[n] := x[i];
		x[i] := x[n-i-1];
		x[n-i-1] := x[n];
	end;
setlength(x, n);
var n,i: integer;
	tmp:TElement;
[...]
n:= length(x);
for i := 0 to (n div 2)-1 do
	begin
		tmp := x[i];
		x[i] := x[n-i-1];
		x[n-i-1] := tmp;
	end;
my @list = ('words', 'of', 'list', 'a', 'reverse');
my @reversed = reverse @list;
reverse(x, Reversed).
y = x[::-1]
x.reverse()
x = reversed(x)
x.reverse!
let y: Vec<_> = x.into_iter().rev().collect();
x.reverse();
x = x.reverse
(reverse x)
X reversed.
Dim ItemList As New List(Of String)(New String() {"one", "two", "three"})
ItemList.Reverse()
For Each item In ItemList
    Console.WriteLine(item)
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