计算x的平方根

 

@Test
    public void test() {
        double res = sqrt(2, 0.00001);
        System.out.println(res);
    }

    /**
     * 求x的平方根
     * 因为x的平方，这个函数是单调递增的，可以考虑使用二分法
     * @param x
     * @param precision 精度
     * @return
     */
    private double sqrt(int x, double precision) {
        if (x == 0 || x == 1) {
            return 1;
        }

        double left = 1, right = x;
        while (left <= right) {
//            double mid = left + (right - left) / 2;
            double mid =  (right + left) / 2;
            if (mid == (x / mid)) {
                return mid;
            }

            if ((right - left) < precision) {
                return right;
            }

            if (mid * mid > x) {
                right = mid;
                continue;
            }

            left = mid;
        }

        return right;
    }