LeetCode 342. Power of Four

Description

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.

Analysis

解法一：用(num & (num - 1)) == 0來確定num是2的冪，由於4的冪中，1只可能位於奇數位（最低位設爲第0位），則有num & 0x55555555 == num。
解法二：思路同上，但使用的反向思維，所有的偶數位應都爲0，即num & 0xAAAAAAAA == 0。
解法三：判斷是否2的冪同上，但判斷是否爲4的冪時，借用了3，即(num - 1) % 3 == 0。

Code

Version 1

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

Version 2

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

Version 3

class Solution {
public:
    bool isPowerOfFour(int num) {
        return ((num & (num - 1)) == 0) && ((num - 1) % 3 == 0);
    }
};

Appendix

• Link: https://leetcode.com/problems/power-of-four/
• Run Time:
  
  • Version 1: 3ms
  • Version 2: 6ms
  • Version 3: 3ms

Question: don’t know why the 2nd solution is slower than others…