0009. Palindrome Number (E)

Palindrome Number (E)

题目

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

Example 1:

Input: 121
Output: true

Example 2:

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

Follow up:

Coud you solve it without converting the integer to a string?

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题意

判断一个整数是否为回文数。

思路

不转换为字符串，直接对整数进行处理，有比较多的方法，如比较逆序后整数与输入的整数是否相同、将整数每一位拆分出来存进数组等。可以对第一个方法进行优化：不需要完全逆序，将逆序一半的整数和剩下的一半比较大小即可。

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代码实现

Java

class Solution {
    public boolean isPalindrome(int x) {
        // 末尾为0且输入不为0是特殊情况，需要先排除
        if (x < 0 || (x != 0 && x % 10 == 0)) {
            return false;
        }
        int reverse  = 0;
        while (x > reverse) {
            reverse = reverse * 10 + x %10;
            x /= 10;
        }
        // 有奇偶两种情况
        return x == reverse || x == reverse / 10;
    }
}

JavaScript

/**
 * @param {number} x
 * @return {boolean}
 */
var isPalindrome = function (x) {
  if (x < 0 || (x !== 0 && x % 10 === 0)) return false

  let y = 0
  while (x > y) {
    y = y * 10 + (x % 10)
    x = Math.trunc(x / 10)
  }

  return x === y || x === Math.trunc(y / 10)
}