feat: solve No.231

BaffinLee · BaffinLee · commit f6a75fad7c5e · 2024-02-19T23:12:57.000+08:00
diff --git a/201-300/231. Power of Two.md b/201-300/231. Power of Two.md
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+# 231. Power of Two
+
+- Difficulty: Easy.
+- Related Topics: Math, Bit Manipulation, Recursion.
+- Similar Questions: Number of 1 Bits, Power of Three, Power of Four.
+
+## Problem
+
+Given an integer `n`, return **`true` if it is a power of two. Otherwise, return `false`**.
+
+An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
+
+ 
+Example 1:
+
+```
+Input: n = 1
+Output: true
+Explanation: 20 = 1
+```
+
+Example 2:
+
+```
+Input: n = 16
+Output: true
+Explanation: 24 = 16
+```
+
+Example 3:
+
+```
+Input: n = 3
+Output: false
+```
+
+ 
+**Constraints:**
+
+
+	
+- `-231 <= n <= 231 - 1`
+
+
+ 
+**Follow up:** Could you solve it without loops/recursion?
+
+## Solution
+
+```javascript
+/**
+ * @param {number} n
+ * @return {boolean}
+ */
+var isPowerOfTwo = function(n) {
+    if (n <= 0) return false;
+    if (n === 1) return true;
+    if (n % 2) return false;
+    return isPowerOfTwo(n / 2);
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(log(n)).
+* Space complexity : O(1).
