LeetCode 9 - Palindrome Number

Difficulty: easy

Problem Description

English (Palindrome Number)

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

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:

• -2^31 <= x <= 2^31 - 1

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

Chinese (回文数)

给你一个整数 x ，如果 x 是一个回文整数，返回 true ；否则，返回 false 。

回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。

• 例如，121 是回文，而 123 不是。

示例 1：

输入：x = 121
输出：true

示例 2：

输入：x = -121
输出：false
解释：从左向右读, 为 -121 。 从右向左读, 为 121- 。因此它不是一个回文数。

示例 3：

输入：x = 10
输出：false
解释：从右向左读, 为 01 。因此它不是一个回文数。

提示：

• -2^31 <= x <= 2^31 - 1

Solution

Notice some edge cases:

• negative number
• zero
• multiples of 10
• the number after palindrome is over the threshold of integer

C++

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        if (x == 0) {
            return true;
        }
        // below is the occasion that x > 0
        if (x % 10 == 0) {
            return false;
        }

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

        return (palindrome_num == x) || (palindrome_num / 10 == x);
    }
};