问题
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?
答案1
0x55555555 is to get rid of those power of 2 but not power of 4
so that the single 1 bit always appears at the odd position
public boolean isPowerOfFour(int num) {
return num > 0 && (num&(num-1)) == 0 && (num & 0x55555555) != 0;
}答案2
(4^n - 1) % 3 == 0
proof:
1. 4^n - 1 = (2^n + 1) * (2^n - 1)
2. among any 3 consecutive numbers, there must be one that is a multiple of 3 among (2^n-1), (2^n), (2^n+1), one of them must be a multiple of 3, and (2^n) cannot be the one, therefore either (2^n-1) or (2^n+1) must be a multiple of 3, and 4^n-1 must be a multiple of 3 as well.
public boolean isPowerOfFour(int num) {
return num > 0 && (num & (num - 1)) == 0 && (num - 1) % 3 == 0;
}