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79 lines (65 loc) · 1.8 KB
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package math;
import org.junit.Assert;
import org.junit.Test;
/**
* 数值的整数次方
* 题目：实现函数double Power(double base, int exponent)，求base的exponent
* 次方。不得使用库函数，同时不需要考虑大数问题。
* Created by hxchen on 2018/6/3.
*/
public class JMath {
public static double power(double base, int exponent) {
double result = 1d;
if (base == 0) {
// 0的正数幂都等于0;负数幂没有定义,但是我们可以在此定义成0
return 0;
} else {
if (exponent == 0) {
return result;
} else if (exponent < 0) {
for (int i = 0; i < -exponent; i++) {
result = base * result;
}
result = 1 / result;
} else if (exponent > 0) {
for (int i = 0; i < exponent; i++) {
result = base * result;
}
}
return result;
}
}
@Test
public void test1() {
Assert.assertEquals(power(0, 1), 0d, 0.000001);
}
@Test
public void test2() {
Assert.assertEquals(power(0, -1), 0d, 0.000001);
}
@Test
public void test3() {
Assert.assertEquals(power(0, 0), 0d, 0.000001);
}
@Test
public void test4() {
Assert.assertEquals(power(2, 3), 8d, 0.000001);
}
@Test
public void test5() {
Assert.assertEquals(power(2, -3), 0.125, 0.000001);
}
@Test
public void test6() {
Assert.assertEquals(power(-2, 3), -8, 0.000001);
}
@Test
public void test7() {
Assert.assertEquals(power(-2, -3), -0.125, 0.000001);
}
@Test
public void test8() {
int code = 3;
Assert.assertTrue(code == 3);
}
}