LeetCode 50 计算pow(x,n),给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。

主要思想是递归，代码如下;

class Solution {
public:
    double Power(double base, int exponent) {
          if(exponent<0){
              return 1/Power(base,-exponent);
          }else if(exponent==0){
                  return 1;
          }else{
              double half=Power(base,exponent/2);
              return exponent%2?base*half*half:half*half;
          }
    }
};