Given an integer (signed 32 bits), write a function to check whether it is a power of 4.
Example:
Given num = 16, return true. Given num = 5, return false.
Follow up: Could you solve it without loops/recursion?
含义:判断一个数是否为4的n次方
方法一: http://blog.csdn.net/yums467/article/details/51425679 //换底公式
1 public boolean isPowerOfFour(int num) { 2 if(num < 1) return false; 3 if(num == 1) return true; 4 int n = (int) (Math.log10(num)/Math.log10(4)); 5 return Math.pow(4,n) == num; 6 }
方法二: num & (num - 1)用来判断num是否是2的次方 4的次方数减一,能被3整除
1 public boolean isPowerOfFour(int num) { 2 if(num < 1) return false; 3 if(num == 1) return true; 4 return num > 0 && (num & (num - 1)) == 0 && (num - 1) % 3 == 0; 5 }