实现pow(x , n)函数。

这里我们看下自带的math.h里是怎么实现的。

//自带实现

template<class _Ty> inline
        _Ty _Pow_int(_Ty _X, int _Y)
        {unsigned int _N;
        if (_Y >= 0)
                _N = (unsigned int)_Y;
        else
                _N = (unsigned int)(-_Y);
        for (_Ty _Z = _Ty(1); ; _X *= _X)
                {if ((_N & 1) != 0)
                        _Z *= _X;
                if ((_N >>= 1) == 0)
                        return (_Y < 0 ? _Ty(1) / _Z : _Z); }}

  double pow(double x, int n) {
     unsigned int N;
     if(n >= 0) N = (unsigned int)n;
     else N = (unsigned int)(-n);
     for(double z = 1; ; x *= x ){
         if((N & 1) != 0)
         z *= x;
         if((N >>= 1) == 0)
         return (n < 0 ? 1/z : z);
     }
    }

看一下discuss里面一位coder的问题：

 double pow(double x, int n) {
        if(n==0) return 1;
        if(n==1) return x;
        if(x==0) return x;
        if(n<0) return pow(1/x, -n);
        if(n%2) return x*pow(x*x, n/2);
        else return pow(x*x, n/2);
    }
Last executed input: 1.00000, -2147483648

However, my code below works! I just change one line! But, I think both are same!

class Solution {
public:
    double pow(double x, int n) {
        if(n==0) return 1;
        if(n==1) return x;
        if(x==0) return x;
        if(n<0){
            x = 1/x;
            n = -n;
        }

        if(n%2) return x*pow(x*x, n/2);
        else return pow(x*x, n/2);
    }
}

if(n<0) return pow(1/x, -n);

Pow(x, n)