文章目录
C++快速幂
题目描述
解题思路
代码
class Solution {
public:
double myPow(double x, int n)
{
return n < 0 ? 1.0 / pow(x, -(long long)n) : pow(x, n);
}
double pow(double x, long long n)
{
if(n == 0) return 1;
double tmp = pow(x, n / 2);
return n % 2 == 0 ? tmp * tmp : tmp * tmp * x;
}
};