Five simple ways to tackle Palindrome Number (Easy) | LeetCode Practice #2

Palindrome Number

Given an integer x, return true if x is a palindrome, and false otherwise.

Python

####Two Pointer (Runtime: 19ms, Memory: 12.3MB)
    #DECLARE x: INTEGER
class Solution:
    def isPalindrome(self, x):
        y = str(x)
        Head = 0
        Tail = len(y) - 1
        while Head < Tail:
            if y[Head] != y[Tail]:
                return False
            Head += 1
            Tail -= 1
        return True



####Recursion (Runtime: 19ms, Memory: 12.3MB)
    #DECLARE x: INTEGER
class Solution:
    def isPalindrome(self, x):
        y = str(x)
        Length = len(y)
        if y[0] != y[-1]:
            return False
        if Length <= 2:
            return True
        return self.isPalindrome(y[1:-1])



####Slicing Radius-Check(Runtime: 0ms, Memory: 12.3MB)
    #DECLARE x: INTEGER
class Solution:
    def isPalindrome(self, x):
        y = str(x)
        Radius = len(y) // 2
        return y[:Radius] == y[::-1][:Radius]



####Mathematical Calculation (Runtime: 27ms, Memory: 12.3MB)
    #DECLARE x: INTEGER
class Solution:
    def isPalindrome(self, x):
        if x < 0:
            return False
        Rev = 0
        Temp = x
        while Temp > 0:
            Rev = Rev * 10 + (Temp%10)
            Temp = Temp // 10
        return x == Rev



####Mathematical Radius-Check (Runtime: 9ms, Memory: 12.4MB)
    #DECLARE x: INTEGER
class Solution:
    def isPalindrome(self, x):
        if x < 0 or ((x%10==0) and (x!=0)):
            return False
        Temp = x
        Rev = 0
        while Temp > Rev:
            Rev = Rev * 10 + (Temp%10)
            Temp = Temp // 10
        if Temp == Rev or Temp == (Rev//10):
            return True
        return False