Update 191._number_of_1_bits.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/191._number_of_1_bits.md

@@ -1,13 +1,34 @@
-### 191. Number of 1 Bits
+# 191. Number of 1 Bits
 
-题目:
+**<font color=red>难度: Easy</font>**
 
-<https://leetcode.com/problems/number-of-1-bits/>
+## 刷题内容
 
+> 原题连接
 
-难度:
+* https://leetcode.com/problems/number-of-1-bits/description/
 
-Easy
+> 内容描述
+
+```
+Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).
+
+Example 1:
+
+Input: 11
+Output: 3
+Explanation: Integer 11 has binary representation 00000000000000000000000000001011 
+Example 2:
+
+Input: 128
+Output: 1
+Explanation: Integer 128 has binary representation 00000000000000000000000010000000
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
 
 
 转成二进制，数1的个数
@@ -22,7 +43,8 @@ class Solution(object):
         return bin(n).count('1')
 ```
 
-
+> 思路 2
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
 
 有wikipedia的题目 [Hamming Weight]((https://zh.wikipedia.org/wiki/汉明重量))
 
@@ -32,13 +54,17 @@ class Solution(object):
 
 原理是在于每次使用x & x-1 总会把低位的数字给置0
 
-比如 3 = 011  2 = 010  3 & 2 = 010 cnt =1
-
-​	2 = 010  1 = 001  2 & 1 = 000 cnt = 2
-
-比如 9 = 1001  8 = 1000  9&8 = 1000 cnt =1
+比如 
+```
+3 = 011  2 = 010  3 & 2 = 010 cnt = 1
+2 = 010  1 = 001  2 & 1 = 000 cnt = 2
+```
 
-​	8 = 1000  7 = 0111  8&7 = 0000 cnt = 2
+比如 
+```
+9 = 1001  8 = 1000  9&8 = 1000 cnt = 1
+8 = 1000  7 = 0111  8&7 = 0000 cnt = 2
+```
 
 > 减1操作将最右边的符号从0变到1，从1变到0，与操作将会移除最右端的1。如果最初X有N个1，那么经过N次这样的迭代运算，X将减到0。下面的算法就是根据这个原理实现的。
 
@@ -59,8 +85,8 @@ class Solution(object):
         """
         cnt = 0
         while n != 0:
-        	n &= n - 1
-        	cnt += 1
-        return cnt 
+            n &= n - 1
+            cnt += 1
+        return cnt == 1
 ```