find the only set bit of a number

Author: ZapDos7

src/binary/rightmostBit/OnlySetBitFinder.java

@@ -0,0 +1,26 @@
+package binary.rightmostBit;
+
+import utils.InputReader;
+
+public class OnlySetBitFinder {
+    public static void main(String[] args) {
+        System.out.print("Input a number: ");
+        int x = InputReader.readInt();
+        System.out.print("Position of only set bit: " + positionOfSet(x));
+    }
+
+    public static int positionOfSet(int n) {
+        // unset the rightmost bit and check if the number is non-zero
+        if ((n & (n - 1)) != 0)
+        {
+            System.out.println("Wrong input");
+            return 1;
+        }
+        // If 0, we can find the position of the only set bit by directly using log2(n) + 1.
+        return log(n, 2) + 1;
+    }
+
+    public static int log(int x, int base) {
+        return (int) (Math.log(x) / Math.log(base));
+    }
+}