Description
Given an integer n, return true if n has exactly three positive divisors. Otherwise, return false.
An integer m is a divisor of n if there exists an integer k such that n = k * m.
Example 1:
Input: n = 2
Output: false
Explantion: 2 has only two divisors: 1 and 2.
Example 2:
Input: n = 4
Output: true
Explantion: 4 has three divisors: 1, 2, and 4.
Constraints:
Solutions
Solution 1
PythonJavaC++GoJavaScript
class Solution:
def isThree(self, n: int) -> bool:
return sum(n % i == 0 for i in range(2, n)) == 1(code-box)
class Solution {
public boolean isThree(int n) {
int cnt = 0;
for (int i = 2; i < n; i++) {
if (n % i == 0) {
++cnt;
}
}
return cnt == 1;
}
}(code-box)
class Solution {
public:
bool isThree(int n) {
int cnt = 0;
for (int i = 2; i < n; ++i) {
cnt += n % i == 0;
}
return cnt == 1;
}
};(code-box)
func isThree(n int) bool {
cnt := 0
for i := 2; i < n; i++ {
if n%i == 0 {
cnt++
}
}
return cnt == 1
}(code-box)
/**
* @param {number} n
* @return {boolean}
*/
var isThree = function (n) {
let cnt = 0;
for (let i = 2; i < n; ++i) {
if (n % i == 0) {
++cnt;
}
}
return cnt == 1;
};(code-box)
Solution 2
PythonJavaC++GoJavaScript
class Solution:
def isThree(self, n: int) -> bool:
cnt = 0
i = 1
while i <= n // i:
if n % i == 0:
cnt += 1 if i == n // i else 2
i += 1
return cnt == 3(code-box)
class Solution {
public boolean isThree(int n) {
int cnt = 0;
for (int i = 1; i <= n / i; ++i) {
if (n % i == 0) {
cnt += n / i == i ? 1 : 2;
}
}
return cnt == 3;
}
}(code-box)
class Solution {
public:
bool isThree(int n) {
int cnt = 0;
for (int i = 1; i <= n / i; ++i) {
if (n % i == 0) {
cnt += n / i == i ? 1 : 2;
}
}
return cnt == 3;
}
};(code-box)
func isThree(n int) bool {
cnt := 0
for i := 1; i <= n/i; i++ {
if n%i == 0 {
if n/i == i {
cnt++
} else {
cnt += 2
}
}
}
return cnt == 3
}(code-box)
/**
* @param {number} n
* @return {boolean}
*/
var isThree = function (n) {
let cnt = 0;
for (let i = 1; i <= n / i; ++i) {
if (n % i == 0) {
cnt += ~~(n / i) == i ? 1 : 2;
}
}
return cnt == 3;
};(code-box)
