力扣算法题—050计算pow（x, n）

#include "000库函数.h"



//使用折半算法  牛逼算法
class Solution {
public:
    double myPow(double x, int n) {
        if (n == 0)return 1;
        double res = 1.0;
        for (int i = n; i != 0; i /= 2) {
            if (i % 2 != 0)
                res *= x;
            x *= x;
        }
        return n > 0 ? res : 1 / res;
    }

};


//同样使用二分法，但是使用递归思想
class Solution {
public:
    double myPow(double x, int n) {
        if (n == 0)return 1;
        double res = myPow(x, n / 2);
        if (n % 2 == 0)return x * x;//是偶数，则对半乘了之后再两个大数相乘，x^n==(x^n/2)^2
        if (n > 0) return res * res * x;
        return res * res / x;
    }
};

void T050() {
    Solution s;
    double x;
    int n;
    x = 0.0001;
    n = 2147483647;
    cout << s.myPow(x, n) << endl;
    x = 2.1;
    n = 3;
    cout << s.myPow(x, n) << endl;
    x = 2;
    n = -2;
    cout << s.myPow(x, n) << endl;
    x = 0.9;
    n = 2147483647;
    cout << s.myPow(x, n) << endl;


}