最小公倍数算法挑战
function smallestCommons(arr) {
var max = Math.max(arr[0], arr[1]);
var min = Math.min(arr[0], arr[1]);
var j = 1, lock;
while (true) {
lock = true;
for (var i = min; i <= max; i++) {
if (j % i != 0) {
lock = false;
break;
}
}
if (lock) {
break;
}
j++;
}
console.log(j);
return j;
}