## 题目描述

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:

 1 2 3  Input: 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits. 

### Example 2:

 1 2 3  Input: 00000000000000000000000010000000 Output: 1 Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit. 

### Example 3:

 1 2 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.

## 解题思路

### n & n - 1

#### 普通正整数

$$N = 2^n*tn + 2^{n-1}t{n-1} + … + 2^0t0$$

$$10 = 2^3 * 1 + 2^2 * 0 + 2^1 * 1 + 2^0 * 0$$

$$9 = 2^3 * 1 + 2^2 * 0 + 2^1 * 0 + 2^0 * 1$$

10 & 9就是将10减去其中的最小的2的幂2，得到的结果为8

N = N & (N-1)会将N的二进制表示中最右边的1变成0。

## 代码

• solution.go
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33  package l166 import ( "log" "strconv" ) func hammingWeight(num uint32) int { if num == 0 { return 0 } var count = 1 for { num &= (num - 1) if num == 0 { break } count += 1 } return count } func binStrToUint32(binary string) uint32 { num, err := strconv.ParseUint(binary, 2, 32) if err != nil { log.Fatalln(err) } return uint32(num) } 
• solution_test.go
  1 2 3 4 5 6 7 8 9 10 11 12 13 14  package l166 import ( "testing" "github.com/stretchr/testify/assert" ) func TestHammingWeight(t *testing.T) { assert.Equal(t, hammingWeight(binStrToUint32("00000000000000000000000000001011")), 3) assert.Equal(t, hammingWeight(binStrToUint32("11111111111111111111111111111101")), 31) assert.Equal(t, hammingWeight(binStrToUint32("00000000000000000000000010000000")), 1) assert.Equal(t, hammingWeight(binStrToUint32("00000000000000000000000000000000")), 0) }