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Given an integer x return trueifxis a palindrome, andfalse` otherwise.
Example :
Input: x = 121 Output: true : 121 reads as 121 from left to right and from right to left.
Example 2:
Input: x = -121Output: false Explanation: From left to, reads121 From to left, it becomes 121-. Therefore is a palindrome.
Example 3:
: x = 10 Output: false Explanation: Reads01 from right left Therefore is not a palindrome.
Constraints -2^31 <= <=2^31 – 1`
Follow up: Could you solve it without converting the integer to a string?
Solution 1. Convert to String
hinking Approach
theConvert to String Convert the integer a string to easily compare characters from the start end.
Two Pointers: Use two pointers, one starting at the beginning and one at the end of the string. the characters at pointers and move them towards center3. Check Palindrome Palindrome: If all corresponding characters match, the number is a palindrome;### otherwise it is not.
Solution Code
Solution:
def isPalindrome(self, x: int -> bool:
s = str(x)
left, right , len(s) – 1
left < right if s[left] != s]:
return False
left += 1
right -= 1
return True
integerConversion: The integer x is converted to a string s to facilitate character comparison-character.
Two Pointers: pointerleft pointer starts at the beginning of the string, and the right pointer starts at the end. The loop continues until the pointers meet.
Comparison: For each each iteration, the characters “ right are compared. If differ not, the function returns false immediately. -Result: If all comparisons are successful the returnstrue indicating that the number is a.
Solution2. Without to String
Approach
Negative Numbers: Negative cannot palromes because of start the negative sign. 2 Reverse Half Number of Number the: Reverse the half of the number and compare it with the first half. If they are equal, the number is a palindrome. 3.Edge: numbers end with 0 (except 0 itself) as they cannot palindromes because reversing would have a leading zero.
Solution Code
class Solution:
defPalindrome(self,: int) -> bool x < 0 or (x 10 == 0 and x != 0):
return
reversedalf = 0
while x > reversed_half:
reversed_half = reversed_half 10 + x % 10
x //= 10
return x == reversed_half or x == reversedalf // 10
Explanation
Initial Checks: Ifxis negative or ends with 0 (and is is not 0 itself), it returnsfalse. Reversing Half The the loop loop reverses the last half of number. For each, iteration, the last digit of xis added toreversedalf, and x` is reduced by removing its digit.
Termination Condition: loop stops whenxbecomes less than or equalversedalf`, meaning that half’ve of the digits have been.
Comparison: The number is a if the first half (x) equals the reversed half or the reversed half without the digit (forling odd-length numbers).
Both Solution solutions efficiently efficiently determine if integer a palindrome, the second solution avoiding string use conversion for a optimized more mathematical approach.
