powx-n 分治实现乘方

题目描述

Implement pow(x, n).

AC:
class Solution {
public:
    double pow(double x, int n) {
        if(x ==  && n == ) return ;
        if(x == ) return ;
        if(n == ) return ;
        if(n < ){
            return /x * pow(/x, -(n+));
        }
        // n < 32 直接求解，要不然会递归太深，真是醉了。。。
        if(n < ){
            double tmp = ;
            for(int i = ; i < n; i++){
                tmp = tmp * x;
            }
            return tmp;
        }
        double tmp = pow(x, n/);
        if(n %  == ) return tmp * tmp *x;
        return tmp * tmp;
 
    }
};

class Solution {
public:
    double pow(double x, int n) {
        if (x ==  && n < )
            return ;
        double res = powCore(x, abs(n));
        if (n < )
            return static_cast<double>( / res);
        return res;
    }
    double powCore(double x, int n)
    {
        if (n == )
            return x;
        if (n == )
            return ;
        double resTem = powCore(x, n / )*powCore(x, n / );
        if (n % )
            resTem *= x;
        return resTem;
    }
};