CODEWITHSUNDEEP
javamath
user2837251
user2837251

Reputation:

How to check in Java if a number is a cube

In Java how do you check if a number is a cube ?

The number can be between the range −2,147,483,648..2,147,483,647

Say for instance given the following numbers we can see that they are cubes

8 (2^3) - True
27 (3^3) - True
64 (4^3) - True

Upvotes: 1

Views: 5738

Answers (5)

Sean Owen
Sean Owen

Reputation: 66886

(-1291)^3 and 1291^3 are both already outside the range of an int in Java. So there are 2581 such numbers anyway. Honestly a lookup table might be the easiest and fastest thing.

Upvotes: 5

user2813148
user2813148

Reputation:

Try some Math (java.lang.Math):

boolean isCube(double input) {
    double cubeRoot = Math.cbrt(input); // get the cube root
    return Math.round(cubeRoot) == cubeRoot; // determine if number is integral
    // Sorry for the stupid integrity determination. I tried to answer fast 
    // and really couldn't remember the better way to do that :)
}

Upvotes: 2

alephreish
alephreish

Reputation: 540

Try to take a cubic root, round the result and take its cube:

int a = (int) Math.round(Math.pow(number_to_test, 1.0/3.0));
return (number_to_test == a * a * a);

Upvotes: 2

Ingo
Ingo

Reputation: 36349

Well, you could do the following (pseudocode)

double x = number;
int b = floor (x ^ (1.0/3.0))  // ^ is exponentiation
if (b*b*b == number || (b+1)*(b+1)*(b+1) == number)
    // it is a cube

Upvotes: 1

Justin Chang
Justin Chang

Reputation: 194

Add a loop:

int inputNum = //whatever ;
int counter = 1;
boolean po3 = false;
while(counter<inputNum || po3==true){
 if((counter*counter*counter)==inputNum){
  po3 = true;
 } else {
  counter++;
 }
}

Upvotes: -2

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