power of 2 - x&(x-1)==0 where x>0

查看多少个1

n == (n&-n) - 利用2’s complement, negating的话可以查看原来的数究竟有没有1

** 前提是n>0

let’s do the 套路

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def poweroffour(n):
	if n==1: return True
	count = 0
	if (n & (n-1))==0: #if power of 2
		while n>1:
			n>>=1 #divided by 2 or shift right
			count+=1
		if count%2==0:
			return True
	
	return False

power of 3 - find max integer of 3’s power and modulo it

len(bin(3**x))-2 < 32, x==19