Add solution 231

Author: yanglbme

README.md

@@ -8,8 +8,6 @@
 
 </center>
 
-# <center>LeetCode Solutions</center>
-
 ## Introduction
 Complete solutions to Leetcode problems, updated daily.
 
@@ -29,6 +27,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 189 | [Rotate Array](https://github.com/yanglbme/leetcode/tree/master/solution/189.Rotate%20Array) | `Array` |
 | 198 | [House Robber](https://github.com/yanglbme/leetcode/tree/master/solution/198.House%20Robber) | `Dynamic Programming` |
 | 203 | [Remove Linked List Elements](https://github.com/yanglbme/leetcode/tree/master/solution/203.Remove%20Linked%20List%20Elements) | `Linked List` |
+| 231 | [Power of Two](https://github.com/yanglbme/leetcode/tree/master/solution/231.Power%20of%20Two) | `Math`, `Bit Manipulation` |
 | 237 | [Delete Node in a Linked List](https://github.com/yanglbme/leetcode/tree/master/solution/237.Delete%20Node%20in%20a%20Linked%20List) | `Linked List` |
 | 344 | [Reverse String](https://github.com/yanglbme/leetcode/tree/master/solution/344.Reverse%20String) | `Two Pointers`, `String` |
 | 876 | [Middle of the Linked List](https://github.com/yanglbme/leetcode/tree/master/solution/876.Middle%20of%20the%20Linked%20List) | `Linked List` |

solution/231.Power of Two/README.md

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+## 2的幂
+### 题目描述
+
+给定一个整数，编写一个函数来判断它是否是 2 的幂次方。
+
+示例 1:
+```
+输入: 1
+输出: true
+解释: 2^0 = 1
+```
+
+示例 2:
+```
+输入: 16
+输出: true
+解释: 2^4 = 16
+```
+
+示例 3:
+```
+输入: 218
+输出: false
+```
+
+### 解法
+可以利用 2^31 对该数取余，结果为 0 则为 2 的幂次方。
+```java
+class Solution {
+    public boolean isPowerOfTwo(int n) {
+        return n > 0 && 1073741824 % n == 0;
+    }
+}
+```
+
+也可以循环取余，每次除以 2，判断最终结果是否为 1。
+```java
+class Solution {
+    public boolean isPowerOfTwo(int n) {
+        if (n < 1) {
+            return false;
+        }
+        
+        while (n % 2 == 0) {
+            n >>= 1;
+        }
+        return n == 1;
+    }
+}
+```

solution/231.Power of Two/Solution.java

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+class Solution {
+    public boolean isPowerOfTwo(int n) {
+        if (n < 1) {
+            return false;
+        }
+        
+        while (n % 2 == 0) {
+            n >>= 1;
+        }
+        return n == 1;
+    }
+}