1 Description #
source: https://leetcode.com/problems/power-of-four/
Given an integer n, return true if it is a power of four. Otherwise, return false.
An integer n is a power of four, if there exists an integer x such that n == 4^x.
Example 1:
Input: n = 16
Output: true
Example 2:
Input: n = 5
Output: false
Example 3:
Input: n = 1
Output: true
Constraints:
-2^31 <= n <= 2^31 - 1
Follow up: Could you solve it without loops/recursion?
2 Solution #
#include <bitset>
class Solution {
public:
bool isPowerOfFour(int n) {
// Space complexity: O(33) -> O(1)
// Time complexity: O(33) -> O(1)
std::bitset<33> binary = std::bitset<33>(n);
if (binary.count() != 1) {
return false;
}
std::string bin_str = binary.to_string();
for (int i = bin_str.size() - 1; i >= 0; i--) {
if (bin_str[i] == '1') {
return (bin_str.size() - 1 - i) % 2 == 0;
}
}
return false;
}
};