Leetcode-9. Palindrome Numbber

subject

Palindrome Number

Given an integer x, return true if x is palindrome integer.

An integer is a palindrome when it reads the same backward as forward.

For example, 121 is a palindrome while 123 is not.

Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Example 2:

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Constraints:

-2³¹ <= x <= 2³¹ - 1

Follow up: Could you solve it without converting the integer to a string?

solution

Method 1 ： String inversion

public class PalindromeNumber9 {
    

    public static void main(String[] args) {
    
        int a = 121;
        boolean palindrome = Solution.isPalindrome(a);
        System.out.println(palindrome);

    }

    /** *  Reverse string  * * @author aiwenwen * @date 11:08 2022/6/10 **/
    static class Solution {
    
        static public boolean isPalindrome(int x) {
    
            String revBefore = String.valueOf(x);

            StringBuffer strUtil = new StringBuffer();
            strUtil.append(revBefore);
            String revAfter = strUtil.reverse().toString();

            if (revBefore.equals(revAfter)) {
    
                return true;
            }

            return false;
        }
    }
}

Method 2 ： Reverse half number

    /** *  Reverse half number  * * @author aiwenwen * @date 11:08 2022/6/10 **/
    static class Solution2 {
    
        static public boolean isPalindrome(int x) {
    
            /** * A special case ： *  As mentioned above , When x<0 when ,x Not palindrome . *  similarly , If the last digit is 0, To make the number palindrome , *  The first digit should also be 0 *  Only 0 Satisfy this attribute  **/
            if (x < 0 || (x % 10 == 0 && x != 0)) {
    
                return false;
            }

            int revertedNumber = 0;
            while (x > revertedNumber) {
    
                revertedNumber = revertedNumber * 10 + x % 10;
                x /= 10;
            }

            /** * * When the number length is odd , We can go through revertedNumber / 10 Remove numbers in the middle . *  for example , When the input is  12321  when , stay while At the end of the loop we can get  x = 12, revertedNumber = 123, *  Because the number in the middle does not affect palindrome （ It is always equal to itself ）, So we can simply remove it . **/
            boolean b = x == revertedNumber;
            boolean b1 = x == revertedNumber / 10;

            return b || b1;

        }
    }

Complexity analysis

Time complexity ：O(logn), For each iteration , We'll divide the input by 10, So the time complexity is zero O(logn).
Spatial complexity ：O(1). We just need constant space to hold a number of variables .

Reprint

author ：LeetCode-Solution
link ：https://leetcode.cn/problems/palindrome-number/solution/hui-wen-shu-by-leetcode-solution/
source ： Power button （LeetCode）
The copyright belongs to the author . Commercial reprint please contact the author for authorization , Non-commercial reprint please indicate the source .