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; }