231. Power of Two 342. Power of Four -- 判断是否为2、4的整数次幂

231. Power of Two

Given an integer, write a function to determine if it is a power of two.

class Solution {

public:

    bool isPowerOfTwo(int n) {

        return n >  ? (n & (n-)) ==  : false;

    }

};

342. Power of Four

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?

    bool isPowerOfFour(int num) {

        static int mask = 0b01010101010101010101010101010101;

        //edge case

        if (num<=) return false;

        // there are multiple bits are 1

        if ((num & num-) != ) return false;

        // check which one bit is zero, if the place is 1 or 3 or 5 or 7 or 9...,

        // then it is the power of 4

        if ((num & mask) != ) return true;

        return false;

    }

巴特西

231. Power of Two 342. Power of Four -- 判断是否为2、4的整数次幂

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