342. Power of Four【位运算】

2017/3/23 22:23:57

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?


思路:题目要求不能循环或递归,这里采用位运算与逻辑运算直接出结果。
 
1、指数运算是必须大于0;
2、首先满足2的指数(只有一个bit是1),利用num&num-1==0即可判断;
3、去掉只满足2的指数不满足4的指数的值,0x2AAAAAAA对应的bit位不能是1。100B = 4 属于4的指数,1000B=8仅仅是2的指数。
 
版本1  Java   O(1)
public class Solution {
    public boolean isPowerOfFour(int num) {
		return num > 0 && ( num & num - 1 | num & 0x2AAAAAAA ) == 0 ;
    }
}

  

原文地址:https://www.cnblogs.com/flyfatty/p/6624810.html