《剑指offer》面试题11 数值的整数次方 Java版

书中方法：这道题要注意底数为0的情况。double类型的相等判断。乘方的递归算法。

	public double power(double base, int exponent){
		//指数为0
      	 if(exponent == 0){
			return 1.0;
		}
      	//底数为0
		if(isEqual(base, 0.0)){
			return 0.0;
		}

		int absExponent = exponent;
		if(exponent < 0)absExponent = -absExponent;

		double result = unsignedPower(base, absExponent);

		if(exponent<0){
			result = 1.0/result;
		}

		return result;
	}
	//求乘方正常思路
	private double unsignedPower(double base, int absExponent){
		double result = 1.0;
		for(int i=1; i<=absExponent; i++){
			result *= base;
		}
		return result;
	}
	//用递归的方法求乘方
	private double unsignedPower2(double base, int absExponent){
		if(absExponent == 1)return base;
		double result = unsignedPower2(base, absExponent>>1);
		result *= result;
		if((absExponent & 0x01) == 1){
			result *= base;
		}
		return result;
	}
	//double类型判断相等
	private boolean isEqual(double num1, double num2){
		if((num1 - num2)>-0.0000001 && (num1 - num2)<0.0000001){
			return true;
		}else{
			return false;
		}
	}