技術標籤：筆試演算法java

題目描述：

程式碼：

public class Design {
    static double PowerWithExponent(double base, int exponent) {
        // base為0的處理
        if (base == 0) {
            if (exponent < 0) {
                return 0;
            } else {
                return 1;
            }
        }
        if (exponent == 0) {
            return 1;
        }
        if (exponent == 1) {
            return base;
        }
        int temp = exponent;
        if (exponent < 0) {
            temp = -temp;
        }
        double result = PowerWithExponent(base, temp >> 1);
        result *= result;
        if ((temp & 1) == 1) {
            result *= base;
        }
        if (exponent < 0) {
            return 1 / result;
        }
        return result;
    }

    static double PowerWithExponentTired(double base, int exponent) {
        // base為0的處理
        if (base == 0) {
            if (exponent < 0) {
                return 0;
            } else {
                return 1;
            }
        }
        if (exponent == 0) {
            return 1;
        }
        if (exponent == 1) {
            return base;
        }
        int temp = exponent;
        if (exponent < 0) {
            temp = -temp;
        }
        double result = PowerWithExponentTired(base, temp >> 1) * PowerWithExponentTired(base, temp >> 1);
        if ((temp & 1) == 1) {
            result *= base;
        }
        if (exponent < 0) {
            return 1 / result;
        }
        return result;
    }

    public static void main(String[] args) throws Exception {
        double base = -2.0;
        for (int i = -100; i < 101; i++) {
            double first = PowerWithExponent(base, i);
            double second = PowerWithExponentTired(base, i);
            if (first != second) {
                System.out.println("bad bad");
            }
        }
    }
}
 

解析：

當為偶數時

當為奇數時

看起來就是個遞迴，對於需要求兩次，如果用遞迴的話，要避免對求兩次，一次遞迴就可以解決的。參考第一個方法，將遞迴算出的平方一下即可，比第二種方法好。