191. Number of 1 Bits #

Problem #

Write a function that takes an unsigned integer and return the number of ‘1’ bits it has (also known as the  Hamming weight).

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using  2’s complement notation. Therefore, in Example 3 above the input represents the signed integer -3.

Problem Summary #

Write a function whose input is an unsigned integer and returns the number of digit bits that are ‘1’ in its binary representation (also known as the Hamming weight).

Solution Approach #

• Find the number of 1s in the binary bits of a uint32 number.
• The approach for this problem is to use binary bit operations. The operation X = X & ( X -1 ) can clear the lowest set binary bit 1. Use this operation until the number is cleared to zero. The number of operations is the number of binary bits that are 1.
• The simplest method is to directly call the library function bits.OnesCount(uint(num)).

Code #

package leetcode

import "math/bits"

// Solution 1
func hammingWeight(num uint32) int {
	return bits.OnesCount(uint(num))
}

// Solution 2
func hammingWeight1(num uint32) int {
	count := 0
	for num != 0 {
		num = num & (num - 1)
		count++
	}
	return count
}