要点：
1 n的正负数
2 double 类型的
class Solution {
public:
double pow(double x, int n) {
if(equal(x,0.0)&&n<0){
return 0;
}
double res=1;
if(n<0){
res=1.0/power(x,-n);
}else{
res=power(x,n);
}
return res;
}
double power(double x,int n){
if(n==0){
return 1;
}
double temp=1;
if(n%2==0){
temp*=power(x*x,n/2);
}else{
temp=temp*x*pow(x*x,n/2);
}
return temp;
}
bool equal(double num1,double num2){
if(fabs(num1-num2)<0.000000001){
return 1;
}
return 0;
}
};