GCD超过2个数字

Question

function gcd (a, b) {
    if(b == 0){
        return a;
    }
    return gcd(b, a%b);
}

function gcd_more_than_two_numbers (a) {
    var last = Math.max.apply(null, a);
    var first = Math.min.apply(null, a);    

    return gcd(first, last);
}

console.log(gcd_more_than_two_numbers([9999,213123,9,15,27])); 
console.log(gcd_more_than_two_numbers([5,10,15,25]));

Is it right to take the lowest and the highest values in an array and find a gcd for all the numbers between them ? Is it mathematically right ?

Answer 1

Is it right to take the lowest and the highest values in an array and find a gcd for all the numbers between them ? Is it mathematically right ?

NO .

---

You need to take the gcd of the first pair and then recalculate against all the other elements of the array, you can do that easily with reduce :

 function gcd (a, b) { if(b == 0){ return a; } return gcd(b, a%b); } function gcd_more_than_two_numbers (a) { return a.reduce(gcd) } console.log(gcd_more_than_two_numbers([9999,213123,9,15,27])) 

Answer 2

No it isn't.

The identity you're after is gcd(a, b, c) = gcd(gcd(a, b), c) .

I suggest you unpick the recursion using a loop.

Answer 3

  function gcd() { var arr = Array.prototype.slice.call(arguments); return arr.reduce(function(a, b) { if (b == 0) { return a; } return gcd(b, a % b); }); } console.log(gcd(7,14)); console.log(gcd(9999,213123,9,15,27)); console.log(gcd(5,10,15,25)); 

Answer 4

You can use array.every as well to check validity:

Sample

 function getGCD(arr) { var min = Math.min.apply(null, arr); var gcd = 1; for (var i = gcd + 1; i <= min; i++) { if (arr.every(x => x % i === 0)) gcd = i; } return gcd; } var arr = [100, 13000, 1110]; var arr2 = [9999, 213123, 9, 15, 27] console.log(getGCD(arr)) console.log(getGCD(arr2))