LC9 Palindrome Number

class Solution:
# time O(N) space O(N)
# main idea: string the number and compare from both end ; edge case: num <0  false 
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        s = str(x)
        l = 0
        r = len(s)-1
        while l < r:
            if s[l] != s[r]:
                return False
            l += 1
            r -= 1
        return True

class Solution:
# time O(lg(N)) spaceO(1)
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        def reverse(x):
            rev = 0
            while x != 0:
                pop = x%10
                x //= 10
                #  be aware of overflow 
                rev = rev*10 + pop
            return rev 
        rev = reverse(x)
        return rev == x

class Solution:
# timeO(lgN)  space O(1)
    # main idea: haf reverse stop point is when x <= reversed bottom half number 
    def isPalindrome(self, x: int) -> bool:
        #edge case: negtive number and lowest digit numbe is 0 
        if x < 0 or (x>0 and x%10==0):
            return False
        rev = 0
        while x > rev:
            rev = x %10 + rev*10
            x//=10
            
        return x == rev or x == rev//10  # odd and even number cases

class Solution:
    # main idea: count digits in number
    def isPalindrome(self, x: int) -> bool:
        #edge case: negtive number and num ==  0 
        if x < 0:
            return False
        if x == 0:
            return True
        digits = math.floor(math.log(x,10) +1) 
        rev = 0
        pop = 0
        for i in range(digits//2):
            pop = x % 10
            rev = pop+ rev*10
            x //= 10
        if digits%2 ==0 and x ==rev :
            return True
        if digits%2 != 0 and x//10 ==rev:
            return True
        return False