一、题目

Given a 32-bit signed integer, reverse digits of an integer.
Example 1：
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note: 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<Integer> 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; }}