50

//快速幂算法（递归）
class Solution {
    public double myPow(double x, int n) {
        if(x == 0){
            return 0;
        }
        long N = n;
        if(N < 0){
            x = 1 / x;
            N = -N;
        }
        return fastPow(x, N);
    }
    public double fastPow(double x, long N){
        if (0 == N){
            return 1.0;
        }
        double half = fastPow(x, N/2);
        if(N % 2 == 0){
            return half * half;
        }
        else{
            return half * half * x;
        }
    }
}