📝 docs : add 191. Number of 1 Bits.md

Author: GoogTech

docs/LeetCod刷题之旅及题目解析/191.位1的个数/191.位1的个数.md

@@ -0,0 +1,47 @@
+## 191. 位1的个数
+> https://leetcode-cn.com/problems/number-of-1-bits/
+
+
+### Java
+```java
+/*
+ * @Author: Goog Tech
+ * @Date: 2020-08-16 10:07:51
+ * @LastEditTime: 2020-08-16 10:08:33
+ * @Description: https://leetcode-cn.com/problems/number-of-1-bits/
+ * @FilePath: \leetcode-googtech\#191. Number of 1 Bits\Solution.java
+ * @WebSite: https://algorithm.show/
+ */
+
+public class Solution {
+    // you need to treat n as an unsigned value
+    public int hammingWeight(int n) {
+        // 使用与(&)运算,即两位同时为1时结果才为1
+        return n == 0 ? 0 : 1 + hammingWeight(n & (n - 1)); 
+    }
+}
+```
+
+### Python
+```python
+'''
+Author: Goog Tech
+Date: 2020-08-16 10:07:58
+LastEditTime: 2020-08-16 10:08:49
+Description: https://leetcode-cn.com/problems/number-of-1-bits/
+FilePath: \leetcode-googtech\#191. Number of 1 Bits\Solution.py
+WebSite: https://algorithm.show/
+'''
+
+class Solution(object):
+    def hammingWeight(self, n):
+        """
+        :type n: int
+        :rtype: int
+        """
+        count = 0
+        # bin() 返回一个整数 int 或者长整数 long int 的二进制表示
+        for i in str(bin(n)):
+            if i == '1': count += 1
+        return count
+```

docs/SUMMARY.md

@@ -48,6 +48,7 @@
     * [160.相交链表](LeetCod刷题之旅及题目解析/160.相交链表/160.相交链表.md)
     * [167.两数之和II-输入有序数组](LeetCod刷题之旅及题目解析/167.两数之和II-输入有序数组/167.两数之和II-输入有序数组.md)
     * [182.查找重复的电子邮箱](LeetCod刷题之旅及题目解析/182.查找重复的电子邮箱/182.查找重复的电子邮箱.md)
+    * [191.位1的个数](LeetCod刷题之旅及题目解析/191.位1的个数/191.位1的个数.md)
     * [203.移除链表元素](LeetCod刷题之旅及题目解析/203.移除链表元素/203.移除链表元素.md)
     * [222.完全二叉树的节点个数](LeetCod刷题之旅及题目解析/222.完全二叉树的节点个数/222.完全二叉树的节点个数.md)
     * [226.翻转二叉树](LeetCod刷题之旅及题目解析/226.翻转二叉树/226.翻转二叉树.md)