LeetCode 50. Pow(x, n)

Posted 2021-01-04 約束の空

暴力会超时，利用二分优化。

class Solution {
public:
    double myPow(double x, int n) {
        if (n<0){
            x = 1/x;
            n = -n;
        }
        return pow(x,n);
    }
    
    double pow(double x, int n){
        if (n==0) return 1.0;
        double half=pow(x,n/2);
        if (n%2==0) return half*half;
        else return half*half*x;
    }
};