[LeetCode] 342. Power of Four（位操作）

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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.

Follow up: Could you solve it without loops/recursion?

思路

题意：不使用循环和递归，求一个数是否是4的幂次。

题解：我们知道凡是4的幂次，那么其二进制位肯定只有一个1其余都是0，并且可以发现规律，从4的零次幂开始列举，可以发现与上一个幂次相比，隔了一个0，也就是将4的幂次的二进制位写在一起，呈现出...0101...的形式，因此根据 num & (num - 1)可以判定二进制位是否只有一个1，然后再与十六进制的55555555相与即可判断是否有4的幂次的表现形式，因为十六进制的5的表现形式正好是0101。

class Solution {
public://3msbool isPowerOfFour(int num) {return !(num & (num - 1)) && (num & 0x55555555);}
};

转载于:https://www.cnblogs.com/ZhaoxiCheung/p/7400404.html

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[LeetCode] 342. Power of Four（位操作）

Description

思路

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