leetcode 34 : Pow(x, n)

class Solution {
public:
double pow(double x, int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

if( n==0) return 1;
if( n==1) return x;

bool sign = n>0 ? true : false;
if(!sign) n = -n;

double rel=0;

rel = pow(x, n/2);
rel *= rel;

if(n%2){
rel *= x;
}

return sign ? rel : 1.0/rel;
}
};

public class Solution {
public double pow(double x, int n) {
// Start typing your Java solution below
// DO NOT write main() function
if(x==1.0) return 1;
boolean negative = false;
if(n<0) {
negative = true;
n=-n;
}
double res = 1.0;

while(n>0) {
if( (n & 1) ==1) {
res *= x;
}
x*=x;
n>>=1;
}
return negative ? 1.0/res : res;
}
}