题目：

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?

思路：

• 题意是判断一个32位的符号整数是不是4的次方
• 对于2的次方的判断是n&（n-1）== 0
  
  10 => 2
  
  100 => 4
  
  1000 => 8
  
  10000 => 16
  
  100000 => 32
  
  1000000 => 64
  
  10000000 => 128
  
  100000000 => 256
  
  1000000000 => 512
  
  10000000000 => 1024
  
  100000000000 => 2048
  
  1000000000000 => 4096
  
  10000000000000 => 8192
  
  100000000000000 => 16384
  
  由图中观察可以看出来，4的次方，1都在从右往左数的奇数位，1，3，5等
  
  所有从2的次方移除4的次方，与上01010101010101010101010101010101，十六进制是0x555555555

代码：

public class Solution {
    public boolean isPowerOfFour(int num) {
        return num > 0 && (num&(num -1)) == 0 && (num & 0x55555555) != 0;
    }
}