231 Power of Two

Author: int32bit

README.md

@@ -125,6 +125,7 @@
 + [217 Contains Duplicate(hashset)](algorithms/ContainsDuplicate)
 + [219 Contains Duplicate II(hashmap)](algorithms/ContainsDuplicateII)
 + [226 Invert Binary Tree(递归，树)](algorithms/InvertBinaryTree)
++ [231 Power of Two(位运算，二进制1的个数)](algorithms/PowerofTwo)
 
 ## Database
 

algorithms/PowerofTwo/README.md

@@ -0,0 +1,30 @@
+## Power of Two
+
+Given an integer, write a function to determine if it is a power of two.
+
+Credits:
+Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
+
+## Solution
+
+位运算，一个数是2的幂，则它的二进制1的个数等于1
+
+求一个数二进制的1的个数的方法是不断地`n &= (n - 1)`，直到n为0
+
+考虑以下两种情况:
+
+* `n == 0`， 此时`n & (n - 1) == 0`, 但不是2的幂
+* `n=-2147483648`，此时二进制1的个数也等于1，但它不是2的幂
+
+综合可以得出：只要`n <= 0`, 就一定不是2的幂。
+
+因此最后解决方案为:
+
+```c
+bool isPowerOfTwo(int n)
+{
+	if (n <= 0)
+		return false;
+	return !(n & (n - 1));
+}
+```

algorithms/PowerofTwo/solve.c

@@ -0,0 +1,17 @@
+#include <stdio.h>
+#include <stdbool.h>
+#include <stdlib.h>
+bool isPowerOfTwo(int n)
+{
+	if (n <= 0)
+		return false;
+	return !(n & (n - 1));
+}
+int main(int argc, char **argv)
+{
+	int n;
+	while (scanf("%d", &n) != EOF) {
+		printf("%d\n", isPowerOfTwo(n));
+	}
+	return 0;
+}