> For the complete documentation index, see [llms.txt](https://mm.ricky.moe/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mm.ricky.moe/algorithm/algorithm-and-data-structure/leetcode/7.-zheng-shu-fan-zhuan.md).

# 7. 整数反转

## 题目

给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。

**示例 1:**

```
输入: 123
输出: 321
```

 **示例 2:**

```
输入: -123
输出: -321
```

**示例 3:**

```
输入: 120
输出: 21
```

**注意:**

假设我们的环境只能存储得下 32 位的有符号整数，则其数值范围为 \[−231,  231 − 1]。请根据这个假设，如果反转后整数溢出那么就返回 0。

## JavaScript实现

1. 转换为string后逆序转换
2. 对于-号进行预处理
3. 对于溢出的数字return前处理

```javascript
/*
 * @lc app=leetcode.cn id=7 lang=javascript
 *
 * [7] 整数反转
 */
/**
 * @param {number} x
 * @return {number}
 */
var reverse = function(x) {
    const s = x.toString()
    let reverseS = ''
    s[0] === '-' ? reverseS+='-': null
    for (let i = s.length-1; i >=0; i--) {
      if(s[i] === '-') continue
      reverseS += s[i]
    }

    const reverseInt = parseInt(reverseS)
    if(reverseInt<-Math.pow(2,31) || reverseInt>Math.pow(2,31)-1){
      return 0
    }else{
      return reverseInt
    }
};


```


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