Update 231. _Power_of_Two.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/231. _Power_of_Two.md

@@ -1,42 +1,63 @@
-### 231.  Power of Two
+# 231. Power of Two
 
+**<font color=red>难度: Easy</font>**
 
+## 刷题内容
 
-题目:
+> 原题连接
 
-<https://leetcode.com/problems/power-of-two/>
+* https://leetcode.com/problems/power-of-two/description/
 
-难度:
+> 内容描述
 
-Easy
+```
+Given an integer, write a function to determine if it is a power of two.
+
+Example 1:
+
+Input: 1
+Output: true 
+Explanation: 20 = 1
+Example 2:
+
+Input: 16
+Output: true
+Explanation: 24 = 16
+Example 3:
 
+Input: 218
+Output: false
+```
 
+## 解题方案
 
-思路：
+> 思路 1
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
 
 
 
-power of two 那是这个数字的binary 表示一定只有一个1
+power of two 那是这个数字的 binary 表示一定只有一个1
 
-套用以前的代码[leetcode191](https://github.com/Lisanaaa/thinking_in_lc/blob/master/191._number_of_1_bits.md)
+套用以前的代码[leetcode191](https://github.com/apachecn/awesome-algorithm/blob/master/docs/Leetcode_Solutions/Python/191._number_of_1_bits.md)
 
 这样会超时
 
 ```
 class Solution(object):
-    def isPowerOfTwo(self, n): # 此法超时
-        """
+    def isPowerOfTwo(self, n):  # 此法超时
+        """
         :type n: int
         :rtype: bool
         """
         cnt = 0
         while n != 0:
-        	n &= n - 1
-        	cnt += 1
+            n &= n - 1
+            cnt += 1
         return cnt == 1
 ```
 
-
+> 思路 2
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
 
 跟power of three一样递归，可以AC
 
@@ -55,14 +76,14 @@ class Solution(object):
             return True
         if n % 2 == 0:
             return self.isPowerOfTwo(n/2)
-        return False
-                	
+        return False            	
 ```
 
 
 
 
-
+> 思路 3
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
 
 
 也是有[算法的wikipedia page](https://en.wikipedia.org/wiki/Power_of_two#Fast_algorithm_to_check_if_a_positive_number_is_a_power_of_two)