leetcode 34 : Pow(x, n)

Implement 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;
    }
}