力扣：50. Pow(x, n)

1、用递归形式的二分快速幂去求解 

class Solution {
public:
    double myPow(double x, int n) {
        if(n<0)
        {
             long long m=-(long long)n;
             return 1/qpow(x,m);
        }
        else  return qpow(x,n);   
    }
    double qpow(double x,long long n)
    {
        if(n==0) return 1;
        double v=myPow(x*x,n/2);
        if(n&1) v=v*x;
        return v;
    }
};

标签：myPow,int,Pow,50,Solution,力扣,double
来源： https://blog.csdn.net/qq_50917103/article/details/121136770