一、题目

Given a 32-bit signed integer, reverse digits of an integer.

Example 1： Input: 123 Output: 321Example 2:  Input: -123 Output: -321Example 3:  Input: 120 Output: 21Note: Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1].For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

二、解题思路：

1、定义一个List集合；

2、定义一个循环，取出x中的每一位数并存入List集合中，当循环执行完时集合中每个元素的顺序已是x的倒序；3、循环遍历集合，用元素乘以相应的位数，得到倒序后的数值；4、判断结果是否越界，如越界则返回0，否则返回结果值。

三、代码实现

public int reverse(int x) {    List
    
      originalList = new ArrayList<>();    double result =  0;    int temp = 0;    while (x != 0) {            temp = x % 10;            originalList.add(temp);            x = x / 10;    }    for (int i = 0; i < originalList.size(); i++) {            result = result + originalList.get(i) * (Math.pow(10, originalList.size() - 1 - i));    }    if (result < Math.pow(-2, 31) || result > Math.pow(2, 31) - 1) {            return 0;    } else {            return (int)result;    }}