LeetCode – Power of Four (Java)

Given an integer (signed 32 bits), write a function to check whether it is a power of 4.

Java Solution 1 - Naive Iteration

public boolean isPowerOfFour(int num) {
    while(num>0){
        if(num==1){
            return true;
        }
 
        if(num%4!=0){
            return false;
        }else{
            num=num/4;
        }
    }
 
    return false;
}

Java Solution 2 - Bit Manipulation

public boolean isPowerOfFour(int num) {
    int count0=0;
    int count1=0;
 
    while(num>0){
        if((num&1)==1){
            count1++;
        }else{
            count0++;
        }
 
        num>>=1;
    }
 
    return count1==1 && (count0%2==0);
}

Java Solution 3 - Math Equation

We can use the following formula to solve this problem without using recursion/iteration.

Power of Four

public boolean isPowerOfFour(int num) {
   if(num==0) return false;
 
   int pow = (int) (Math.log(num) / Math.log(4));
   if(num==Math.pow(4, pow)){
       return true;
   }else{
       return false;
   }
}

Category >> Algorithms

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mg

public boolean isPowerOfFour(int n) { if(n > 1; }
return (count0 % 2 == 0); }
Prateek Gupta

why so complex :O ….checkout this one – with explanation
https://discuss.leetcode.com/topic/76414/one-line-java-no-complex-coding-using-only-integer-class-methods

return Integer.bitCount(num) == 1 && (Integer.toBinaryString(num).length()-1)%2==0;
Feiyu Zhou

Here’s a simple solution:
return num > 0 && (num & (num – 1)) == 0 && (num & 0x55555555) == num;
Andrew

In the “num==Math.pow(4, pow)”, it can be replaced by just check whether pow is an integer.

public class Solution {
public boolean isPowerOfFour(int num) {
if(num == 0) return false;

double pow = Math.log(num)/Math.log(4);

if((int) pow == pow){
return true;
}

return false;
}
}

LeetCode – Power of Four (Java)

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