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?

Credits:
Special thanks to @yukuairoyfor adding this problem and creating all test cases.

给一个有符号的32位整数，写一个函数检查此数是否为4的次方数。

解法1：位操作

解法2：循环

解法3: 数学函数, 换底公式

Java:

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:

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:

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;

   }

}

Python:

class Solution(object):

    def isPowerOfFour(self, num):

        """

        :type num: int

        :rtype: bool

        """

        return num > 0 and (num & (num - 1)) == 0 and \

               ((num & 0b01010101010101010101010101010101) == num)

Python:

# Time:  O(1)

# Space: O(1)

class Solution2(object):

    def isPowerOfFour(self, num):

        """

        :type num: int

        :rtype: bool

        """

        while num and not (num & 0b11):

            num >>= 2

        return (num == 1)

C++:

class Solution {

public:

    bool isPowerOfFour(int num) {

        while (num && (num % 4 == 0)) {

            num /= 4;

        }

        return num == 1;

    }

};

C++:

class Solution {

public:

    bool isPowerOfFour(int num) {

        return num > 0 && int(log10(num) / log10(4)) - log10(num) / log10(4) == 0;

    }

};

C++:

class Solution {

public:

    bool isPowerOfFour(int num) {

        return num > 0 && !(num & (num - 1)) && (num & 0x55555555) == num;

    }

};

C++:　　

class Solution {

public:

    bool isPowerOfFour(int num) {

        return num > 0 && !(num & (num - 1)) && (num - 1) % 3 == 0;

    }

};

类似题目：

[LeetCode] 231. Power of Two 2的次方数

[LeetCode] 326. Power of Three 3的次方数

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巴特西

[LeetCode] 342. Power of Four 4的次方数

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