Leetcode 9. palindromes

Power button | #9 Palindrome number

Title screenshot

Method 1 ： Convert an integer to a string, invert the characters, and then compare

Learn from the idea of integer inversion , Slightly modify the code , Add a returned comparison value

class Solution:
    def isPalindrome(self, x: int) -> bool:
        INT_MIN, INT_MAX = -2**31, 2**31-1
        if x<INT_MIN or x>INT_MAX:
            return False
        else:
            y = int(str(abs(x))[::-1])

        if y<INT_MIN or y > INT_MAX:
            return False
        else:
            return x==y

Code optimization

Directly compare the two converted strings , There is no need to switch back and forth .

class Solution:
    def isPalindrome(self, x: int) -> bool:
        INT_MIN, INT_MAX = -2**31, 2**31-1
        if x<INT_MIN or x>INT_MAX:
            return False
        else:
            x = str(x)
            return x == x[::-1]

Code re optimization

class Solution:
    def isPalindrome(self, x: int) -> bool:
        INT_MIN, INT_MAX = -2**31, 2**31-1
        s=str(x)
        return True if (x>INT_MIN and x<INT_MAX) and (s == s[::-1]) else False

Method 2 ： Mathematical method

Negative numbers must not be palindromes , Go straight back to False. Positive integer pair 10 After taking the module, it becomes the head of the new number . Comparing whether the old and new numbers are the same .

class Solution:
    def isPalindrome(self, x: int) -> bool:
        INT_MIN, INT_MAX = -2**31, 2**31-1
        y = 0
        temp = x
        if x < 0:
            return False
        while temp != 0:
            y = y*10 + temp%10
            temp//=10
        return False if  (y>INT_MAX or x>INT_MAX or y<INT_MIN or x<INT_MIN ) else (x == y)

Palindromes can also be considered because they are symmetrical , In fact, you don't need to compare after reversing all , But separate from the middle , before and after comparison .

I don't want to write any more here .