Given an integer (signed 32 bits), write a function to check whether it is a power of 4.
给定一个32位的整数，判断是否为4次幂

　　一开始看到这题我就想到打表，但是，果然还是位运算最厉害

Example:

Given num = 16, return true. Given num = 5, return false.

Follow up

Could you solve it without loops/recursion?

My Solution

(Java) Version 1 Time: 2ms:

　　没有什么好说的，在32位数的世界里面，4次幂并没有几个，只有15个数，直接存数组里面对比就是了，只是没有想到还是慢了

public class Solution {
    public boolean isPowerOfFour(int num) {
        int[] integer = new int[]{1,4,16,64,256,1024,4096,16384,65536,262144,1048576,4194304,16777216,67108864,268435456,1073741824};
        for(int i=0;i<16;i++){
            if(num==integer[i])return true;
        }
        return false;
    }
}

(Java) Version 2 Time: 3ms (By k'):

　　不仅简洁还快，果然所有和2次幂扯上关系的，都可以让位运算来一脚

　　老外的解释：
　　The mask is 01010101010101010101010101010101 (0x55555555), reason for that is any number that is power of 4 happens bit and other bit.

　　For example for 8 bits:
　　4 ^ 0 = 00000001 (first bit is set)
　　4 ^ 1 = 00000100 ( third bit is set)
　　4 ^ 2 = 00010000 (5th bit is set)
　　and so on

　　So the mask would be 0x55555555 (this will take care of the sign bit too)
　　The other part is that you want to make sure that only one bit is set in the number (similar to numbers of power 2), this is done by num == (num & -num)
　　num == (num & -num) will return true only if there was one bit set, the reason has to do about 2's complement representation.

　　For example:
　　4 = 00000100
　　-4 = 11111011 + 1 = 11111100
　　so 00000100 & 11111100 = 00000100

　　The only exception for this rule is if num is 0 which explains why we check num != 0
　　This solution would work for any number that is power of 2 (4, 8, 16, ..), you just need to adjust the mask

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

(Java) Version 3 Time: 21ms (By yfcheng):

　　有意思的从字符串出发的思路，虽然慢就慢了点
　　老外的解释：

The idea is that numbers in quaternary system that is power of 4 will be like 10, 100, 1000 and such. Similar to binary case. And this can be extended to any radix.

public class Solution {
    public boolean isPowerOfFour(int num) {
        return Integer.toString(num, 4).matches("10*");
    }
}

(Java) Version 4 Time: ms (By LeeLom):

　　似乎是一个通用的解法？

public class Solution { 
    public boolean isPowerOfFour(int num) { 
        int x = (int)Math.sqrt(num); 
        //1073741824 is 4^15, 4^16 is bigger than int  
        return(num > 0 && 1073741824 % num == 0 && x*x == num); 
    }
}