Power(base, exponent)by Isai Damier, Android Engineer @ Google

```/*********************************************************************
* Author: Isai Damier
* Title: Power(base, exponent)
* Project: geekviewpoint
* Package: algorithms
*
* Statement: Compute base x raised to exponent n: x^n
*
* Sample Input: 3^11
* Sample Output: 177147
*
* Time Complexity of Solution:
*   Best = Average = Worst = O(Time) = O(log n),
*
* Technical Details:
*   Instead of multiplying x by itself n times using a for-loop,
*   we continually divide the exponent by two (think binary search)
*   hence reducing the time complexity to O(log n) instead of O(n).
*   We demonstrate by manually solving for 3^11.
*
*   3^11 = 3^5 * 3^5 * 3
*   3^5 = 3^2 * 3^2 * 3
*   3^2 = 3 * 3 = 9
*
*  Therefore, we perform five multiplications instead of eleven:
*  3*3=9; 9*9*3 = 243; 243*243*3 = 177147
*
********************************************************************/
public int pow(int base, int exponent) {
if (1 == exponent) {
return base;
}
int result = pow(base, exponent / 2);
if (0 == (exponent & 1)) {
return result * result;
}
return result * result * base;
}```
```import org.junit.Test;
import static org.junit.Assert.*;

public class NumbersTest {

/**
* Test of pow method, of class Numbers.
*/
@Test
public void testPow() {
System.out.println("pow");
Numbers exponent = new Numbers();
int x = 2;
int n = 11;
assertEquals((int) Math.pow(x, n), exponent.pow(x, n));
x = 3;
n = 7;
assertEquals((int) Math.pow(x, n), exponent.pow(x, n));
x = 5;
n = 4;
assertEquals((int) Math.pow(x, n), exponent.pow(x, n));
x = 4;
n = 6;
assertEquals((int) Math.pow(x, n), exponent.pow(x, n));
}
}
```