Leetcode50. Pow(x, n)

Title transmission address : https://leetcode.cn/problems/powx-n/

The code is as follows :

// Solve problems with bit operations 
 public static double myPow(double x, int n) {
    
        //13=8+4+1 1101
        double result = 1;
        double carry = x;
        boolean sign = true;
        long abVal=n;  // The reason to use long, Because n May be int The minimum value of a type -2147483648, In this case, the opposite number will overflow 
        if (n < 0) {
    
            abVal = -abVal;
            sign = false;
        }
        while (abVal > 0) {
    
            int i = (int) (abVal & 1);
            abVal = abVal >> 1;
            if (i == 1) {
    
                result = result * carry;
            }
            carry = carry * carry;
        }
        if (!sign) {
    
            result = (1 / result);
        }
        return result;
    }