【剑指Offer 11】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。

Java代码如下：

package demo;

public class Test10 {
    public static double power(double base, int exponent) {
        if(base == 0 & exponent == 0) {
            throw new RuntimeException("invalid input. base and exponent are both zero!");
        }
        if(base == 0 & exponent < 0) {
            return 0.0;
        }
        // 求幂次方
        double result = powerWithUnsignedExponent(base, Math.abs(exponent));
        if(exponent < 0) {
            result = 1 / result;
        }
        return result;
    }

    /**
     * 求正整数次幂
     * @param base
     * @param exponent
     * @return
     */
    private static double powerWithUnsignedExponent(double base, int exponent) {
        if(exponent == 0) {
            return 1;
        }
        if(exponent == 1) {
            return base;
        }
        // 递归求一半的值
        double result = powerWithUnsignedExponent(base, exponent >> 2);
        result *= result;
        if(exponent % 2 != 0) {
            result *= base;
        }
        return result;
    }

    public static void main(String[] args) {
        // 运行结果为false，说明可以直接用 == 表示2个小数是否相等
        System.out.println(0.00000000000000000000000000000000000000000000001 == 0.0);
        // 运行结果为true，说明写成0或者0.0都可以
        System.out.println(0 == 0.0);
        System.out.println(power(2, 4));
        System.out.println(power(2, -4));
        System.out.println(power(-2, 4));
        System.out.println(power(-2, -4));
        System.out.println(power(2, 0));
        System.out.println(power(0, -4));
        //  System.out.println(power(0, 0));
        System.out.println(power(0.0000000000000000000000000000000000000001, 0));
    }
}

运行结果

来源：http://blog.csdn.net/derrantcm/article/details/45529087