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;

 }