commit solution 191

Author: lxlxw

index-tags.md

@@ -147,6 +147,7 @@
 |  [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>  |✅ |
 |  [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>  |✅ |
 |  [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>  |✅ |
+|  [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>  |✅ |
 |  [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>  |✅ |
 
 #### **树**

index-type.md

@@ -58,7 +58,7 @@
 |  [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>  |
 |  [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>  |✅ |
 |  [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>  |
-|  [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>  |
+|  [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>  |✅ |
 |  [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>  |
 |  [195](https://leetcode-cn.com/problems/tenth-line)  |  [第十行](/solution/100-199/0195.tenth-line/)  |    |  <font color=green>简单</font>  |
 |  [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>  |

solution/100-199/0191.number-of-1-bits/README.md

@@ -0,0 +1,64 @@
+# [191. 位1的个数](https://leetcode-cn.com/problems/number-of-1-bits/description/)
+
+### 题目描述
+
+<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>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>00000000000000000000000000001011
+<strong>输出：</strong>3
+<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>
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>00000000000000000000000010000000
+<strong>输出：</strong>1
+<strong>解释：</strong>输入的二进制串 <strong>00000000000000000000000010000000</strong>&nbsp;中，共有一位为 &#39;1&#39;。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre><strong>输入：</strong>11111111111111111111111111111101
+<strong>输出：</strong>31
+<strong>解释：</strong>输入的二进制串 <strong>11111111111111111111111111111101</strong> 中，共有 31 位为 &#39;1&#39;。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<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>
+	<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>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>进阶</strong>:<br>
+如果多次调用这个函数，你将如何优化你的算法？</p>
+
+### 解题思路
+
+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
+
+![](http://lc-photo.xwlin.com/191.png)
+
+### 具体解法
+
+
+#### **Golang**
+```go
+func hammingWeight(num uint32) int {
+	res := 0
+	for num != 0 {
+		num = num & (num - 1)
+		res++
+	}
+	return res
+}
+```
+
+

solution/100-199/0191.number-of-1-bits/solution.go

@@ -0,0 +1,19 @@
+package leetcode
+
+/*
+ * @lc app=leetcode.cn id=191 lang=golang
+ *
+ * [191] 位1的个数
+ */
+
+// @lc code=start
+func hammingWeight(num uint32) int {
+	res := 0
+	for num != 0 {
+		num = num & (num - 1)
+		res++
+	}
+	return res
+}
+
+// @lc code=end

solution/100-199/0191.number-of-1-bits/solution_test.go

@@ -0,0 +1,17 @@
+package leetcode
+
+import (
+	"testing"
+)
+
+func TestHammingWeight(t *testing.T) {
+	var num uint32
+	var ret int
+
+	num = 00000000000000000000000000001011
+	ret = 3
+
+	if ret != hammingWeight(num) {
+		t.Fatalf("case fails %v\n", ret)
+	}
+}