Reverse Bits/ Number of 1 Bits

Author: gavinfish

190 Reverse Bits.py

@@ -0,0 +1,24 @@
+'''
+Reverse bits of a given 32 bits unsigned integer.
+
+For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
+
+Follow up:
+If this function is called many times, how would you optimize it?
+'''
+
+class Solution(object):
+    def reverseBits(self, n):
+        """
+        :type n: int
+        :rtype: int
+        """
+        n = ((n & 0x55555555) << 1) | ((n >> 1) & 0x55555555)
+        n = ((n & 0x33333333) << 2) | ((n >> 2) & 0x33333333)
+        n = ((n & 0x0f0f0f0f) << 4) | ((n >> 4) & 0x0f0f0f0f)
+        n = (n << 24) | ((n & 0xff00) << 8) | ((n >> 8) & 0xff00) | (n >> 24)
+        return n & 0xffffffff
+
+
+if __name__ == "__main__":
+    assert Solution().reverseBits(43261596) == 964176192

191 Number of 1 Bits.py

@@ -0,0 +1,23 @@
+'''
+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.
+'''
+
+class Solution(object):
+    def hammingWeight(self, n):
+        """
+        :type n: int
+        :rtype: int
+        """
+        n -= (n >> 1) & 0x55555555
+        n = (n & 0x33333333) + ((n >> 2) & 0x33333333)
+        n = (n + (n >> 4)) & 0x0f0f0f0f
+        n += (n >> 8)
+        n += (n >> 16)
+        return n & 0x3f
+
+
+if __name__ == "__main__":
+    assert Solution().hammingWeight(4294967295) == 32
+    assert Solution().hammingWeight(11) == 3