Description
Given an integer n, return the smallest prime palindrome greater than or equal to n.
An integer is prime if it has exactly two divisors: 1 and itself. Note that 1 is not a prime number.
- For example,
2, 3, 5, 7, 11, and 13 are all primes.
An integer is a palindrome if it reads the same from left to right as it does from right to left.
- For example,
101 and 12321 are palindromes.
The test cases are generated so that the answer always exists and is in the range [2, 2 * 108].
Example 1:
Input: n = 6
Output: 7
Example 2:
Input: n = 8
Output: 11
Example 3:
Input: n = 13
Output: 101
Constraints:
Solutions
Solution 1
PythonJavaC++Go
class Solution:
def primePalindrome(self, n: int) -> int:
def is_prime(x):
if x < 2:
return False
v = 2
while v * v <= x:
if x % v == 0:
return False
v += 1
return True
def reverse(x):
res = 0
while x:
res = res * 10 + x % 10
x //= 10
return res
while 1:
if reverse(n) == n and is_prime(n):
return n
if 10**7 < n < 10**8:
n = 10**8
n += 1(code-box)
class Solution {
public int primePalindrome(int n) {
while (true) {
if (reverse(n) == n && isPrime(n)) {
return n;
}
if (n > 10000000 && n < 100000000) {
n = 100000000;
}
++n;
}
}
private boolean isPrime(int x) {
if (x < 2) {
return false;
}
for (int v = 2; v * v <= x; ++v) {
if (x % v == 0) {
return false;
}
}
return true;
}
private int reverse(int x) {
int res = 0;
while (x != 0) {
res = res * 10 + x % 10;
x /= 10;
}
return res;
}
}(code-box)
class Solution {
public:
int primePalindrome(int n) {
while (1) {
if (reverse(n) == n && isPrime(n)) return n;
if (n > 10000000 && n < 100000000) n = 100000000;
++n;
}
}
bool isPrime(int x) {
if (x < 2) return false;
for (int v = 2; v * v <= x; ++v)
if (x % v == 0)
return false;
return true;
}
int reverse(int x) {
int res = 0;
while (x) {
res = res * 10 + x % 10;
x /= 10;
}
return res;
}
};(code-box)
func primePalindrome(n int) int {
isPrime := func(x int) bool {
if x < 2 {
return false
}
for v := 2; v*v <= x; v++ {
if x%v == 0 {
return false
}
}
return true
}
reverse := func(x int) int {
res := 0
for x != 0 {
res = res*10 + x%10
x /= 10
}
return res
}
for {
if reverse(n) == n && isPrime(n) {
return n
}
if n > 10000000 && n < 100000000 {
n = 100000000
}
n++
}
}(code-box)
