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#231. Power of Two ##Question Given an integer, write a function to determine if it is a power of two.

##Solution ###Approach #1 (count the number of 1) [Accepted] ####Algorithm Integer is a power of two means only one bit of n is ‘1’, for example, 100 is 2^2=4 while 110 is 2^2+2^1=6.

When n<=0, it can’t be power of two as 2^-1=0.5 and because the parameter n is int, we can sure that it has only 32 bit, so we count the number of 1 in 32 bits to check whether only one bit of n is ‘1’ when n is positive.

class Solution {
public:
    bool isPowerOfTwo(int n) {
        if(n<=0)
            return false;
        int nums_of_one = 0;
        for(int i=0; i<32; ++i) {
            nums_of_one += n&1;
            if(nums_of_one > 1)
                return false;
            n >>= 1;
        }
        return true;
    }
};

####Complexity Analysis

• Time complexity : O(1).
• Space complexity : O(1).

###Approach #2 (log2 in C++11) [Accepted] ####Algorithm

The result of log2(n) in math must be an interger instead of float when integer is a power of two, so we use log2() function in C++11 and check whether log2(n) is an interger by difference between floor(log2(n)) and ceil(log2(n)).

For example, n=5, log2(5)=2.19722, floor(2.19722)=2, ceil(2.19722)=3, the difference is 1, so it is not power of two.

class Solution {
public:
    bool isPowerOfTwo(int n) {
        if(n<=0)
            return false;

        double tmp = log2(n);
        int a = floor(tmp);
        int b = ceil(tmp);
        if(b-a == 0)
            return true;
        return false;
    }
};

####Complexity Analysis

• Time complexity : O(1).
• Space complexity : O(1).

We can also use pow(2, log2(n)) instead of floor(log2(n)) - ceil(log2(n)).

class Solution {
public:
    bool isPowerOfTwo(int n) {
        if(!n)
            return false;

        int t = floor(log2(n));
        if(pow(2, t) == n)
            return true;
        return false;
    }
};

###Approach #3 (using n&(n-1) trick) [Accepted] ####Algorithm I didn’t come up with this, thanks for dong.wang.1694’s solution.

We can know that power of 2 means only one bit of n is ‘1’, for example, 1, 10, 100 etc, so n-1 means the other bits will become 1, e.g. 0, 01, 011. Power of 2 minus 1 means all of its digits will negate.
Therefore, the result of n&(n-1) must be 0.

class Solution {
public:
    bool isPowerOfTwo(int n) {
        if(n<=0) return false;
        return !(n&(n-1));
    }
};

####Complexity Analysis

• Time complexity : O(1).
• Space complexity : O(1).