Added number of bits

Author: FreeTymeKiyan

Easy/NumberOfBits.java

@@ -0,0 +1,52 @@
+/**
+ * Write a function that takes an unsigned integer and returns the number of
+ * ’1' bits it has (also known as the Hamming weight).
+ * 
+ * For example, the 32-bit integer '11' has binary representation
+ * 00000000000000000000000000001011, so the function should return 3.
+ * 
+ * Tags: Bit Manipulation
+ */
+class NumberOfBits {
+    public static void main(String[] args) {
+        NumberOfBits nob = new NumberOfBits();
+        int n = 111;
+        System.out.println(nob.hammingWeight(n));
+        System.out.println(nob.hammingWeightB(n));
+        System.out.println(nob.hammingWeightC(n));
+    }
+    
+    /**
+     * Pure bit manipulation
+     * "n &= n - 1" is used to delete the right "1" of n
+     * Stop when all 1s are deleted and n is zero
+     */
+    public int hammingWeight(int n) {
+        int res = 0;
+        while (n != 0) {
+            res++;
+            n &= n - 1;
+        }
+        return res;
+    }
+    
+    /**
+     * Most straight forward solution
+     * Iterate 32 times to check each digit in n
+     */
+    public int hammingWeightB(int n) {
+        int res = 0;
+        for (int i = 0; i < 32; i++) // no need to iterate 32 times
+            if ((n >>> i & 0x1) == 1) res++;
+        return res;
+    }
+    
+    /**
+     * Recursive
+     * If n is 0 or 1, return n
+     * If not, return n & 1 + hammingWeightC(n >>> 1)
+     */
+    public int hammingWeightC(int n) {
+        return n == 0 || n == 1 ? n : (n & 1) + hammingWeightC(n>>>1);
+    }
+}