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.
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思路:
(1)该题为给定一个整数,判断该整数是否为4的次方数。
(2)该题属于简单题。用一个循环进行判断即可得到结果。由于用循环比较简单,这里不再累赘。下文简单阐述两种非循环的实现方式,都是基于2的次方数实现的,感兴趣可以看看。
(3)算法代码实现如下,希望对你有所帮助。
package leetcode; public class Power_of_Four { public static boolean isPowerOfFour(int num) { if (num <= 0) return false; if (num == 1) return true; while (true) { if (num % 4 == 0) { num = num / 4; if (num == 1) return true; } else { return false; } } }
网上的一些其它算法实现:
(1)如果该整数是2的次方数,只要是4的次方数,减1之后可以被3整除,则说明该整数是4的次方数,见下方代码:
public class Solution { public boolean isPowerOfFour(int num) { return num > 0 && ((num & (num - 1))==0) && ((num - 1) % 3 == 0); } }
(2)2的次方数在二进制下只有最高位是1,但不一定是4的次方数,观察发现4的次方数的最高位的1都是计数位,只需与上0x55555555(等价 1010101010101010101010101010101),如果得到的数还是其本身,则可以肯定其为4的次方数,见下方代码:
public class Solution { public boolean isPowerOfFour(int num) { return num > 0 && ((num & (num - 1))==0) && (num & 0x55555555) == num; } }