Sword refers to offer 16. The integer power of the value

Sword refers to offer 16. The integer power of the value

Title description

Problem-solving ideas

The difficulty of this question is to consider various boundary conditions.

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Note, why use the long type for exponents of fast powers?

Because the range of int is [-2^31,2^31-1], the range of negative numbers is larger than that of positive numbers, so you cannot simply take the absolute value. If the base is exactly -2^31 and the exponent is of type int, it will overflow.

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To compare double variables for equality, do not use "==", or use the equals method to convert to a string, or the two double variables are reduced to one error (such as 1e-6) as here.

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class Solution {
    
    
    public double myPow(double x, int n) {
    
    
        long N = n;
        //底数为0，指数小于0，则非法
        if (isEqual(x, 0.0) && N < 0) return 0.0;
        //底数为0，指数大于等于0，直接返回0.0
        if (isEqual(x, 0.0) && N >= 0) return 0.0;
        
        return N >= 0 ? quickPow(x, N) : 1.0 / quickPow(x, -N);
        
    }

    //快速幂模板， 注意指数 exponent 是 long 类型防止溢出，且为非负数
    public double quickPow(double base, long exponent) {
    
    
        if (exponent == 0) return 1;
        if (exponent == 1) return base;

        double res = quickPow(base, exponent >> 1);
        res *= res;
        //如果是奇数，则要额外乘上base
        if ((exponent & 0x01) == 1)
            res *= base;
        return res;
    }

    //判断两个 double 变量是否相等
    public boolean isEqual(double a, double b) {
    
    
        return Math.abs(a - b) < 1e-6;
    }
}