题目： 

 

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解析： 

 

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代码： 

public double myPow(double x, int n) {
        return n < 0 ? 1.0 / pow(x,-n) : pow(x,n); 
    }

    private double pow(double x, int n){
        if(n == 0) return 1.0;
        double tmp = pow(x,n / 2);
        return n % 2 == 0 ? tmp * tmp : tmp * tmp * x; 
    }