# [Palindrome Number][title]

## Description

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?

**Tags:** Math


## 思路 0

题意是判断一个有符号整型数是否是回文，也就是逆序过来的整数和原整数相同，首先负数肯定不是，接下来我们分析一下最普通的解法，就是直接算出他的回文数，然后和给定值比较即可。

```java
class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) return false;
        int copyX = x, reverse = 0;
        while (copyX > 0) {
            reverse = reverse * 10 + copyX % 10;
            copyX /= 10;
        }
        return x == reverse;
    }
}
```

## 思路 1

好好思考下是否需要计算整个长度，比如 1234321，其实不然，我们只需要计算一半的长度即可，就是在计算过程中的那个逆序数比不断除 10 的数大就结束计算即可，但是这也带来了另一个问题，比如 10 的倍数的数 10010，它也会返回 `true`，所以我们需要对 10 的倍数的数再加个判断即可，代码如下所示。

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


## 结语

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]



[title]: https://leetcode.com/problems/palindrome-number
[ajl]: https://github.com/Blankj/awesome-java-leetcode