342. Power of Four

题目

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?

链接

https://leetcode.com/problems/power-of-four/?tab=Description

3/8/2017

没有想到follow up的方法

 1 public class Solution {
 2     public boolean isPowerOfFour(int num) {
 3         if (num <= 0) return false;
 4         while (num != 1) {
 5             if (num % 4 != 0) return false;
 6             num /= 4;
 7         }
 8         return true;
 9     }
10 }

他人的好算法

1     public boolean isPowerOfFour(int num) {
2         return num > 0 && (num&(num-1)) == 0 && (num & 0x55555555) != 0;
3         //0x55555555 is to get rid of those power of 2 but not power of 4
4         //so that the single 1 bit always appears at the odd position 
5     }

另外一个，可以用于任何power的例子，只需要把底数改了即可。

1 public boolean isPowerOfFour(int num) {
2     return Integer.toString(num, 4).matches("10*");
3 }

 

posted @ 2017-03-09 01:58  panini  阅读(115)  评论(0)  编辑  收藏  举报