Leetcode50. Pow(x, n)

题目传送地址: ​​https://leetcode.cn/problems/powx-n/​​

代码如下:

//用位运算来解题
 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;  //之所以用long，是因为n可能是int类型的最小值-2147483648，这样的话相反数就会溢出了
        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;
    }