Add solution of SquareRoot task using Babylon method

Author: andrei-punko

README.md

@@ -249,6 +249,7 @@ they contain enough code which describes implementation in a natural way.
 | Java интервью 11                                                                                 | [Youtube](https://youtu.be/zufbVgdBCAI) | -                                                                                                                                                                                   |
 | Java интервью 12                                                                                 | [Youtube](https://youtu.be/PD41epg_pT4) | -                                                                                                                                                                                   |
 | Функциональные тесты REST API с помощью Spock                                                    | [Youtube](https://youtu.be/GK5y3oA3qfM) | -                                                                                                                                                                                   |
+| Вычисление квадратного корня вавилонским методом (leetcode)                                      | [Youtube](https://youtu.be/41zAzebmOuc) | [Code](src/main/java/by/andd3dfx/numeric/SquareRootBabylon.java)                                                                                                                    |
 
 ## Materials & notes
 

src/main/java/by/andd3dfx/numeric/SquareRootBabylon.java

@@ -0,0 +1,37 @@
+package by.andd3dfx.numeric;
+
+/**
+ * <pre>
+ * <a href="https://leetcode.com/problems/sqrtx/description/">Task description</a>
+ *
+ * Given a non-negative integer x, return the square root of x rounded down to the nearest integer.
+ * The returned integer should be non-negative as well.
+ * You must not use any built-in exponent function or operator.
+ *     For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.
+ *
+ * Example 1:
+ *   Input: x = 4
+ *   Output: 2
+ *   Explanation: The square root of 4 is 2, so we return 2.
+ *
+ * Example 2:
+ *   Input: x = 8
+ *   Output: 2
+ *   Explanation: The square root of 8 is 2.82842...,
+ *   and since we round it down to the nearest integer, 2 is returned.
+ * </pre>
+ *
+ * @see <a href="https://youtu.be/41zAzebmOuc">Video solution</a>
+ */
+public class SquareRootBabylon {
+
+    public static int mySqrt(int x) {
+        double old = x;
+        double root = x;
+        do {
+            old = root;
+            root = 0.5 * (root + x / root);
+        } while (Math.abs(root - old) >= 1);
+        return (int) root;
+    }
+}

src/test/java/by/andd3dfx/numeric/SquareRootBabylonTest.java

@@ -0,0 +1,22 @@
+package by.andd3dfx.numeric;
+
+import org.junit.Test;
+
+import static org.assertj.core.api.Assertions.assertThat;
+
+public class SquareRootBabylonTest {
+
+    @Test
+    public void mySqrt() {
+        assertThat(SquareRootBabylon.mySqrt(1)).isEqualTo(1);
+        assertThat(SquareRootBabylon.mySqrt(2)).isEqualTo(1);
+        assertThat(SquareRootBabylon.mySqrt(3)).isEqualTo(1);
+        assertThat(SquareRootBabylon.mySqrt(4)).isEqualTo(2);
+        assertThat(SquareRootBabylon.mySqrt(5)).isEqualTo(2);
+        assertThat(SquareRootBabylon.mySqrt(8)).isEqualTo(2);
+        assertThat(SquareRootBabylon.mySqrt(9)).isEqualTo(3);
+        assertThat(SquareRootBabylon.mySqrt(624)).isEqualTo(24);
+        assertThat(SquareRootBabylon.mySqrt(625)).isEqualTo(25);
+        assertThat(SquareRootBabylon.mySqrt(789)).isEqualTo(28);
+    }
+}