Return the first N items of the Fibonacci sequence.
The input will be a whole number greater than or equal to zero.
The output should be iterable.
Examples:
(0) => [0]
(5) => [0, 1, 1, 2, 3, 5]
(9) => [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
*^You can assume that you will always recieved valid input
Language: PHP
Length: 60
Solution: (Work in progress: as it misses F0 and F1, and becomes inaccurate quite quickly due to the low-precision, you could replace 0.618 with something like ((1 + sqrt(5)) / 2) - 1) for slightly better accuracy.)
function f($n){yield$f=1;for(;$n--;)yield$f=round($f/.618);}Language: Ruby
Length: 44
Solution: (Another work in progress, this one is more accurate than the PHP version)
s=[0];n=k=1;1.upto($*[0].to_i){s<<n;k=n+n=k}
Language: Javascript ES6
Length: 70
Solution: (Produces a full array output as expected, and values of n under 0 produce an empty array.)
f=(n)=>n<0?[]:(a=f(n-1),n==0?[0]:n==1?[0,1]:a.concat([a[n-1]+a[n-2]]))
Language: Javascript ES6
Length: 62
Solution: Further optimisation
f=(n)=>n<2?[0,1].splice(0,n+1):(a=f(n-1),a[n]=a[n-1]+a[n-2],a)
Language: Javascript ES6
Length: 59
Solution: Further optimisation - Correct for all values of n>=0. (Behaves as previous example for all negative values of n except for n=-2, but this annoyance is not required by the problem)
f=n=>n<2?[0,1].slice(0,n+1):(a=f(n-1),a[n]=a[n-1]+a[n-2],a)
Trying a recursive solution for javascript. Wonder if I can do something like [1..n] to generate an array?
Language: Javascript
Length: 69
Solution: