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劍指offer之數值的整數次方

數值的整數次方

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/**
 * @program:
 * @description: 數值的整數次方
 * @author: zhouzhixiang
 * @create: 2018-10-31 20:48
 */
public class Pow {

    public static void main(String[] args) throws Exception {
        System.out.println(power(2,30));
    }

    public static double power(double base, int exponent)
throws Exception { if(equal(base,0.0)&&exponent<0) throw new Exception("無意義"); if(exponent==0) return 1; double result = 0.0; if(exponent<0) result = powerWithExponent(1.0/base,-exponent); else result = powerWithExponent(base,exponent); return result;
} private static double powerWithExponent(double base, int exponent) { double result = 1.0; for(int i = 0; i < exponent; i++){ result *= base; } return result; } private static boolean equal(double base, double v) { return Math.
abs(base-v)<0.00000001 ? true:false; } }