Description#
Given an integer n
, return true
if it is a power of three. Otherwise, return false
.
An integer n
is a power of three, if there exists an integer x
such that n == 3x
.
Example 1:
Input: n = 27
Output: true
Explanation: 27 = 33
Example 2:
Input: n = 0
Output: false
Explanation: There is no x where 3x = 0.
Example 3:
Input: n = -1
Output: false
Explanation: There is no x where 3x = (-1).
Constraints:
Follow up: Could you solve it without loops/recursion?
Solutions#
Solution 1#
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| class Solution:
def isPowerOfThree(self, n: int) -> bool:
while n > 2:
if n % 3:
return False
n //= 3
return n == 1
|
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| class Solution {
public boolean isPowerOfThree(int n) {
while (n > 2) {
if (n % 3 != 0) {
return false;
}
n /= 3;
}
return n == 1;
}
}
|
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| class Solution {
public:
bool isPowerOfThree(int n) {
while (n > 2) {
if (n % 3) {
return false;
}
n /= 3;
}
return n == 1;
}
};
|
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| func isPowerOfThree(n int) bool {
for n > 2 {
if n%3 != 0 {
return false
}
n /= 3
}
return n == 1
}
|
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| function isPowerOfThree(n: number): boolean {
return n > 0 && 1162261467 % n == 0;
}
|
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| /**
* @param {number} n
* @return {boolean}
*/
var isPowerOfThree = function (n) {
return n > 0 && 1162261467 % n == 0;
};
|
Solution 2#
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| class Solution:
def isPowerOfThree(self, n: int) -> bool:
return n > 0 and 1162261467 % n == 0
|
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| class Solution {
public boolean isPowerOfThree(int n) {
return n > 0 && 1162261467 % n == 0;
}
}
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| class Solution {
public:
bool isPowerOfThree(int n) {
return n > 0 && 1162261467 % n == 0;
}
};
|
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| func isPowerOfThree(n int) bool {
return n > 0 && 1162261467%n == 0
}
|