0009 - Palindrome Number (Easy)
Problem Link
https://leetcode.com/problems/palindrome-number/
Problem Statement
Given an integer x, return true if x is palindrome integer.
An integer is a palindrome when it reads the same backward as forward.
- For example,
121is a palindrome while123is not.
Example 1:
Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.
Example 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Constraints:
-2^31 <= x <= 2^31 - 1
Approach 1: Convert to string
Convert the integer value to string format and compare it one by one.
- Python
- Java
- JavaScript
- C++
class Solution:
def isPalindrome(self, x: int) -> bool:
#str: convert to string
#[::-1]: read from the end
return str(x) == str(x)[::-1]
class Solution {
public boolean isPalindrome(int x) {
String str = Integer.toString(x);
StringBuilder rev = new StringBuilder(str).reverse();
return rev.toString().equals(str);
}
}
/**
* @param {number} x
* @return {boolean}
*/
var isPalindrome = function(x) {
var str = x.toString();
return str.split("").reverse().join("") == x.toString();
};
class Solution {
public:
bool isPalindrome(int x) {
string strX = to_string(x);
reverse(strX.begin(), strX.end());
return to_string(x) == strX;
}
};
Approach 2: Calculation
Calculate the reversed number.
In JavaScript, we have used Object.is(..) for value comparison, It's a ES6 feature and makes value comparisons in an even more strict fashion than the === comparison.
References from MDN and You Don't Know JS (Behind the scenes polyfills of Object.is).
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/is
https://github.com/getify/You-Dont-Know-JS/blob/1st-ed/es6%20%26%20beyond/ch1.md#shimspolyfills
- Python
- Java
- JavaScript
- C++
class Solution:
def isPalindrome(self, x: int) -> bool:
in_num = x
reverser = 0
while in_num > 0:
reverser = reverser * 10 + checker % 10
in_num = in_num//10
return reverser == x
class Solution {
public boolean isPalindrome(int x) {
if (x < 0) {
return false;
} else {
int result = 0;
int n = x;
while (n > 0) {
int r = n % 10;
result = result * 10 + r;
n /= 10;
}
return result == x;
}
}
}
var isPalindrome = function(x) {
if (x < 0 || x > 0 && Object.is(x % 10, 0)) return false;
let n = x;
let num = 0;
while (x > 0) {
num = (num * 10) + (x % 10);
x = Math.floor(x / 10);
}
return Object.is(n, num);
};
class Solution {
public:
bool isPalindrome(int x) {
long int res = 0, carry, n = x;
if (x < 0) {
return false;
} else {
while (n > 0) {
carry = n % 10;
res = res * 10 + carry;
n /= 10;
}
}
return res == x;
}
};