题解 | #数值的整数次方#

注意负数的情况，将基数和次方处理一下

这里使用分治减少一半复杂度

class Solution {
public:
    double Power(double base, int exponent) {
      if (exponent < 0) {
        base = 1 / base;
        exponent = -exponent;
      }
      
      double res = 1;
      
      for (int i = 1; i <= exponent / 2; ++i) {
        res *= base;
      }
      
      res *= res;
      
      if (exponent % 2) {
        res *= base;
      }
      
      return res;
    }
};