9. Palindrome Number

四刷 07/2022

Version #1 Reverse the whole number - slower

Time O(logX)

Space O(1)

Runtime: 18 ms, faster than 28.91% of Java online submissions for Palindrome Number.

Memory Usage: 44.6 MB, less than 53.23% of Java online submissions for Palindrome Number.

class Solution {

    public boolean isPalindrome(int x) {

        if (x < 0) {

            return false;

        }

        int original = x;

        long reversed = 0l;

        while (x > 0) {

            reversed = reversed * 10 + x % 10;

            x /= 10;

        }

        return reversed == original;

    }

}

Version #2 Reverse half of the number

写了bug就是当数是10的倍数的时候也是一定是false的

Time O(logX)

Space O(1)

Runtime: 19 ms, faster than 24.56% of Java online submissions for Palindrome Number.

Memory Usage: 44.6 MB, less than 57.89% of Java online submissions for Palindrome Number.

class Solution {

    public boolean isPalindrome(int x) {

        if (x < 0 || x % 10 == 0 && x != 0) {

            return false;

        }

        int reversed = 0;

        while (x > reversed) {

            reversed = reversed * 10 + x % 10;

            x /= 10;

        }

        return reversed == x || reversed / 10 == x;

    }

}

三刷

70.40 % 
class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0 || (x != 0 && x % 10 == 0)) {
            return false;
        }
        int reversed = 0;
        while (reversed < x) {
            reversed = reversed * 10 + x % 10;
            x = x / 10;
        }
        return reversed == x || reversed / 10 == x;
    }

}

一刷 //Compare half of the digits in x, so don't need to deal with overflow.
There are 2 cases, number of digits is an odd number or even number

The number can't be a multiple of ten.
 43.90%
public class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0 || x != 0 && x % 10 == 0) return false;
        int reverse = 0;
        while (x > reverse) {
            reverse = reverse * 10 + x % 10;
            x = x / 10;
        }
        return reverse == x || x == reverse / 10;
    }
} 

二刷 6/3/2018
beats 34.61 %
class Solution {
    public boolean isPalindrome(int x) {
        if (x == 0) return true;
        if (x < 0 || x % 10 == 0) return false;
        int reverse = 0;
        // 121 split from n / 2 + 1, reverse the left half of the number
        // 1, 21
        // origin = 312, reverse = 31
        // if reverse == origin || reverse == origin / 10 -> true
        
        while (reverse < x) {
            reverse = reverse * 10 + x % 10; // 1
            x = x / 10; // 12
        }
        return reverse == x || x == reverse / 10;
    }

}