342. Power of Four

Total Accepted: 4769  Total Submissions: 14338  Difficulty: Easy

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.

Subscribe to see which companies asked this question

Hide Tags

  Bit Manipulation

Show Similar Problems

分析：

老题目了！分析略。

方法1：

反复相除。时间复杂度o(lg(n))

class Solution {
public:
    bool isPowerOfFour(int num) {
        while(num>0 && num%4==0)  
            num/=4;  
        return num==1; 
    }
};

方法2：

log函数取对数

class Solution {
public:
    bool isPowerOfFour(int num) {
        double res = log10(num) / log10(4);  //有精度问题，不要用以指数2.718为低的log函数  
        return (res - int(res) == 0) ? true : false; 
    }
};

方法3：

位运算，没有方法1快呢！

class Solution {
public:
    bool isPowerOfFour(int num) {
        if (num <= 0)
            return false;
        //(num & (num-1)) == 0这个条件有可能是2的幂产生的。
        //(num & 0x55555555) != 0这个条件为的是铲除2的幂
        if ( (num & (num-1)) == 0 && (num & 0x55555555) != 0 )
            return true;

        return false;
    }
};

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/51212846

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895