[LeetCode] Pow(x, n) (二分法)

Implement pow(x, n).

刚开始没想到，后来看remlost的博客才写出来，代码很简练：

class Solution {
public:
    double pow(double x, int n) {
        if(n<0)
            return 1/power(x,-n);
        else
            return power(x,n);
    }
private:
    double power(double x, int n){
        if(n==0)
            return 1;

        double v = power(x,n/2);
        if(n%2==0)
            return v*v;
        else
            return v*v*x;

    }

};

[LeetCode] Pow(x, n) (二分法)