231. Power of Two

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

S1: brutal force O(logn)

    bool isPowerOfTwo(int n) {
        if (n < 1) return false;
        long p = 1;
        while (p < n) {
            p *= 2;
        }
        return p == n;
    }

S2: bit manipulation O(1)

若一个数为2的power，则其二进制表示只有1位为1

    bool isPowerOfTwo(int n) {
        if (n < 1) return false;
        return !(n & (n - 1));
    }

326. Power of Three

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

S1: loop

    bool isPowerOfThree(int n) {
        if (n < 1) return false;
        while(n % 3 == 0) {
            n = n / 3;
        }
        return n == 1;
    }

S2: hard code

max power of 3 that fits in int is 3^19 = 1162261467

    bool isPowerOfThree(int n) {
        return n > 0 && 1162261467 % n == 0;
    }