273. Integer to English Words

Task:

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?

Some Questions:

Can I use the log function?

Solution:

Obviously, we can use loop to check whether the given number is equal to 4,4^2,4^3,..,4^15. 

But how to solve it without loops? In fact, we have solve the problem power of 2. It calculates whether the given number is a divisor of (1<<30).

So, this problem can we solve it by checking whether the given number is a divisor of (4^15)? Obviously, this is not enough, because 2 is also satisfies this condition. Weadditionally have  to ensure it is a even power of 2. How to check it?  One way is to use log function. Let x=log(num)/log(2), if x is a even number than num is power of 4.

Another way is that let x=num|0x55555555, if x==0x55555555 than num is power of 4, because 5 is 0101 in binary system, it filters the number that the most significant digit in odd position.

Code:

class Solution {public:    bool isPowerOfFour(int num) {        if(num<=0)return false;        int HAT=1<<30;        if(HAT%num!=0)return false;        return (num|0x55555555)==0x55555555;    }};

class Solution {public:    bool isPowerOfFour(int num) {        if(num<=0)return false;        int HAT=1<<30;        if(HAT%num!=0)return false;        double x=log(num*1.0)/log(2.0);        int ix=x;        return ix%2==0;    }};