leetcode解题报告233——Power of Two

Given an integer, write a function to determine if it is a power of two.

题目属于easy。前两种解法是一个意思，很容易想到；第三种解法最好（没想到）。

代码：

package bitmanipulation;

//Given an integer, write a function to determine if it is a power of two.

public class PowerofTwo {

public boolean isPowerOfTwo(int n) {
if (n <= 0) {
return false;
}
String s = Integer.toBinaryString(n);
if (s.charAt(0) == '0') {
return false;
} else {
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) == '1') {
return false;
}
}
}
return true;
}

public boolean isPowerOfTwo2(int n) {
if (n <= 0) {
return false;
}
while (n > 0) {
if (n != 1 && n % 2 == 1) {
return false;
}
n = n >> 1;
}
return true;
}

public boolean isPowerOfTwo3(int n) {
if (n <= 0) {
return false;
}
return (n & (n - 1)) == 0;
}

/**
* @param args
*/
public static void main(String[] args) {
System.out.println(new PowerofTwo().isPowerOfTwo3(16));
}

}