# 9. Palindrome Number
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
Input: 121
Output: true
Example 2:
Input: -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: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:
Coud you solve it without converting the integer to a string?
# Solution
Approach 1: Compare the number with its reverse.
Approach 2: Reverse the last half of the number and see if it's equal to the first half of the number.
Approach 3: Starting from both the head and tail of the number, compare each pair of digits towards each other.
# Code (Python)
# Code (C++)
Approach 1:
class Solution {
public:
bool isPalindrome(int x) {
int xCopy = x;
long xReversed = 0;
while (xCopy > 0)
{
xReversed = xReversed * 10 + xCopy % 10;
xCopy /= 10;
}
return x == xReversed;
}
};
Approach 2:
class Solution {
public:
bool isPalindrome(int x) {
// Special case for this approach:
// need to check if the last digit is 0 (e.g., 12210).
if (x < 0 || (x % 10 == 0 && x != 0))
return false;
int x1stHalf = x;
int x2ndHalfReversed = 0;
while (x1stHalf > x2ndHalfReversed)
{
x2ndHalfReversed = x2ndHalfReversed * 10 + x1stHalf % 10;
x1stHalf /= 10;
}
return (x1stHalf == x2ndHalfReversed ||
x1stHalf == x2ndHalfReversed / 10);
}
};
Approach 3:
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0) return false;
int magnitude = 1;
while (x / 10 >= magnitude)
{ // Avoid "x >= magnitude * 10" in case of overflow.
magnitude *= 10;
}
while (magnitude > 1)
{
int head = (x / magnitude) % 10;
int tail = x % 10;
if (head != tail)
return false;
x /= 10;
magnitude /= 100;
}
return true;
}
};