[LeetCode] 342. Power of Four 4的次方数

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 @yukuairoy for 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:

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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:

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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:

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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:

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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:

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# 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++:

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class Solution { public:     bool isPowerOfFour(int num) {         while (num && (num % 4 == 0)) {             num /= 4;         }         return num == 1;     } };

C++:

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class Solution { public:     bool isPowerOfFour(int num) {         return num > 0 && int(log10(num) / log10(4)) - log10(num) / log10(4) == 0;     } };

C++:

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

C++:　　

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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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