Calculate pow(x,n) | GeeksforGeeks

Above function can be optimized to O(logn) by calculating power(x, y/2) only once and storing it.

int power(int x, unsigned int y)

{

    int temp;

    if( y == 0)

        return 1;

    temp = power(x, y/2);

    if (y%2 == 0)

        return temp*temp;

    else

        return x*temp*temp;

}

Let us extend the pow function to work for negative y and float x.
From http://www.programcreek.com/2012/12/leetcode-powx-n/

public double power(double x, int n) {
 if (n == 0)
  return 1;
 
 double v = power(x, n / 2);
 
 if (n % 2 == 0) {
  return v * v;
 } else {
  return v * v * x;
 }
}
 
public double pow(double x, int n) {
 if (n < 0) {
  return 1 / power(x, -n);
 } else {
  return power(x, n);
 }
}

float power(float x, int y)

{

    float temp;

    if( y == 0)

       return 1;

    temp = power(x, y/2);      

    if (y%2 == 0)

        return temp*temp;

    else

    {

        if(y > 0)

            return x*temp*temp;

        else

            return (temp*temp)/x;

    }

}

Also refer to http://www.programcreek.com/2012/12/leetcode-powx-n/
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