剑指offer：数值的整数次方

方法一：

class Solution {
public:
double Power(double base, int exponent) {
if(base==0 && exponent<0){
throw new std::logic_error("error input");
}
if(exponent==0){
return 1;
}

if(exponent==1){
return base;
}

bool minus=false;
if(exponent<0){
minus=true;
exponent=-exponent;
}

double res=1.0;

while(exponent>0){
if(exponent & 1){
res*=base;
}
base*=base;
exponent>>=1;
}

if(minus){
res=1/res;
}

return res;
}
};

方法二：

class Solution {
public:
double Power(double base, int exponent) {

if(exponent==-1){
return 1/base;
}

if(exponent==0){
return 1;
}

if(exponent==1){
return base;
}

double res=Power(base,exponent>>1);
res*=res;

if(exponent & 0x1==1){
res*=base;
}

return res;
}
};