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Commit 25ba5a3

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‎index-tags.md

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| [136](https://leetcode-cn.com/problems/single-number) | [只出现一次的数字](/solution/100-199/0136.single-number/) | `位运算`,`哈希表` | <font color=green>简单</font> |✅ |
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| [169](https://leetcode-cn.com/problems/majority-element) | [多数元素](/solution/100-199/0169.majority-element/) | `位运算`,`数组`,`分治算法` | <font color=green>简单</font> |✅ |
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| [187](https://leetcode-cn.com/problems/repeated-dna-sequences) | [重复的dna序列](/solution/100-199/0187.repeated-dna-sequences/) | `位运算`,`哈希表` | <font color=blue>中等</font> |✅ |
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| [191](https://leetcode-cn.com/problems/number-of-1-bits) | [位1的个数](/solution/100-199/0191.number-of-1-bits/) | `位运算` | <font color=green>简单</font> |✅ |
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| [268](https://leetcode-cn.com/problems/missing-number) | [缺失数字](/solution/200-299/0268.missing-number/) | `位运算`,`数组`,`数学` | <font color=green>简单</font> |✅ |
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#### **树**

‎index-type.md

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| [183](https://leetcode-cn.com/problems/customers-who-never-order) | [从不订购的客户](/solution/100-199/0183.customers-who-never-order/) | | <font color=green>简单</font> |
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| [189](https://leetcode-cn.com/problems/rotate-array) | [旋转数组](/solution/100-199/0189.rotate-array/) | `数组` | <font color=green>简单</font> |✅ |
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| [190](https://leetcode-cn.com/problems/reverse-bits) | [颠倒二进制位](/solution/100-199/0190.reverse-bits/) | `位运算` | <font color=green>简单</font> |
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| [191](https://leetcode-cn.com/problems/number-of-1-bits) | [位1的个数](/solution/100-199/0191.number-of-1-bits/) | `位运算` | <font color=green>简单</font> |
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| [191](https://leetcode-cn.com/problems/number-of-1-bits) | [位1的个数](/solution/100-199/0191.number-of-1-bits/) | `位运算` | <font color=green>简单</font> |✅ |
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| [193](https://leetcode-cn.com/problems/valid-phone-numbers) | [有效电话号码](/solution/100-199/0193.valid-phone-numbers/) | | <font color=green>简单</font> |
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| [195](https://leetcode-cn.com/problems/tenth-line) | [第十行](/solution/100-199/0195.tenth-line/) | | <font color=green>简单</font> |
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| [196](https://leetcode-cn.com/problems/delete-duplicate-emails) | [删除重复的电子邮箱](/solution/100-199/0196.delete-duplicate-emails/) | | <font color=green>简单</font> |
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# [191. 位1的个数](https://leetcode-cn.com/problems/number-of-1-bits/description/)
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### 题目描述
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<p>编写一个函数,输入是一个无符号整数,返回其二进制表达式中数字位数为 &lsquo;1&rsquo;&nbsp;的个数(也被称为<a href="https://baike.baidu.com/item/%E6%B1%89%E6%98%8E%E9%87%8D%E9%87%8F" target="_blank">汉明重量</a>)。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>00000000000000000000000000001011
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<strong>输出:</strong>3
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<strong>解释:</strong>输入的二进制串 <code><strong>00000000000000000000000000001011</strong>&nbsp;中,共有三位为 &#39;1&#39;。</code>
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>00000000000000000000000010000000
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<strong>输出:</strong>1
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<strong>解释:</strong>输入的二进制串 <strong>00000000000000000000000010000000</strong>&nbsp;中,共有一位为 &#39;1&#39;。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入:</strong>11111111111111111111111111111101
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<strong>输出:</strong>31
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<strong>解释:</strong>输入的二进制串 <strong>11111111111111111111111111111101</strong> 中,共有 31 位为 &#39;1&#39;。</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li>请注意,在某些语言(如 Java)中,没有无符号整数类型。在这种情况下,输入和输出都将被指定为有符号整数类型,并且不应影响您的实现,因为无论整数是有符号的还是无符号的,其内部的二进制表示形式都是相同的。</li>
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<li>在 Java 中,编译器使用<a href="https://baike.baidu.com/item/二进制补码/5295284" target="_blank">二进制补码</a>记法来表示有符号整数。因此,在上面的&nbsp;<strong>示例 3</strong>&nbsp;中,输入表示有符号整数 <code>-3</code>。</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>进阶</strong>:<br>
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如果多次调用这个函数,你将如何优化你的算法?</p>
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### 解题思路
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1. num & (num - 1) 作用是消除数字 n 的二进制表示中的最后一个 1
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![](http://lc-photo.xwlin.com/191.png)
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### 具体解法
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#### **Golang**
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```go
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func hammingWeight(num uint32) int {
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res := 0
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for num != 0 {
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num = num & (num - 1)
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res++
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}
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return res
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}
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```
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package leetcode
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/*
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* @lc app=leetcode.cn id=191 lang=golang
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*
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* [191] 位1的个数
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*/
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// @lc code=start
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func hammingWeight(num uint32) int {
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res := 0
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for num != 0 {
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num = num & (num - 1)
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res++
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}
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return res
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}
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// @lc code=end
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package leetcode
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import (
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"testing"
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)
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func TestHammingWeight(t *testing.T) {
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var num uint32
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var ret int
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num = 00000000000000000000000000001011
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ret = 3
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if ret != hammingWeight(num) {
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t.Fatalf("case fails %v\n", ret)
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}
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}

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