解决方案

递归 快速幂

 public double myPow(double x, int n) {
        long N=n;
        if(N<0){
            x= 1/x;
            N=-N;
        }
        return fastPow(x,N);
    }
    private double fastPow(double x ,long n){
        if(n==0)
            return 1.0;
        double half = fastPow(x, n/2);
        if(n%2==0)
            return half*half;
        else
            return half*half*x;
    }