Merge pull request #1 from surajpattanaik/master

Added another approach using bit manipulation

Author: NirmalSilwal

Leetcode/Leetcode_August_Challenge/Day4_PowerOfFour.java

@@ -1,14 +1,17 @@
 package Leetcode_August_Challenge;
-
+import java.util.Scanner;
 public class Day4_PowerOfFour {
 
 public static void main(String[] args) {
-
-int num = 64;
+ Scanner sc = new Scanner(System.in);
+ System.out.println("Enter number");
+int num = sc.nextInt();
 System.out.println(isPowerOfFour(num));
 System.out.println(isPowerOfFour(5));
 System.out.println(isPowerOfFour_log(num));
 System.out.println(isPowerOfFour_log(5));
+System.out.println(isPowerOfFour2(num));
+System.out.println(isPowerOfFour2(5));
 }
 
 // approach 1- using loop
@@ -38,4 +41,26 @@ public static boolean isPowerOfFour_log(int num) {
 // System.out.println(Math.floor(logn) + " "+ Math.ceil(logn));
 return Math.floor(logn) == Math.ceil(logn);
 }
+
+// approach 3
+static boolean isPowerOfFour2(int num) {
+int count = 0;
+
+// Check if there is only one bit set in num
+ int x = num & (num - 1);
+
+if (num > 0 && x == 0) {
+// count 0 bits before set bit
+while (num > 1) {
+num >>= 1;
+count += 1;
+}
+
+// If count is even then return true else false
+return (count % 2 == 0) ? true : false;
+}
+
+// If there are more than 1 bit set then num is not a power of 4
+return false;
+} 
 }