I'm studying recursion and I wrote this method to calculate the N° number of the Fibonacci series:
fibonacci(int n, Map memo) {
if (memo.containsKey(n)) return memo[n]; // Memo check
if (n <= 2) return 1; // base case
// calculation
memo[n] = fibonacci(n - 1, memo) + fibonacci((n - 2), memo);
return memo[n];
}
I think it doesn't need to be explained, my problem is just how to call this function from the main, avoiding providing an empty Map.
this is how I call the function now:
fibonacci(n, {});
But I would rather prefer to call it just like this:
fibonacci(n);
The canonical approach is to make memo
optional, and use a fresh map if the memo
argument is omitted. Because you want to change and update the map, you can't use a default value for the parameter, because default values must be constant, and constant maps are not mutable.
So, written very concisely:
int fibonacci(int n, [Map<int, int>? memo]) {
if (n <= 2) return 1;
return (memo ??= {})[n] ??= fibonacci(n - 1, memo) + fibonacci(n - 2, memo);
}
The ??=
operator assigns to the right-hand side if the value is null
.
It's used both to initialize memo
to a new map if the argument was omitted,
and to update the map if a cached value wasn't present.
I'd actually reconsider using a map. We know that the Fibonacci computation will compute a value for every prior number down to 1, so I'd just use a list instead:
int fibonacci(int n, [List<int?>? memo]) {
if (n <= 2) return 1;
return (memo ??= List<int?>.filled(n - 2))[n - 3] ??=
fibonacci(n - 1, memo) + fibonacci(n - 2, memo);
}
That should work just like the map.
(I subtract 3 from n
when doing the lookup because no value below 3 needs the list - it's handled by the prior if
).