Reverse Integer之Java實現

一、題目

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、解題思路：

1、定義一個List集合；
2、定義一個循環，取出x中的每一位數並存入List集合中，當循環執行完時集合中每個元素的順序已是x的倒序；
3、循環遍歷集合，用元素乘以相應的位數，得到倒序後的數值；
4、判斷結果是否越界，如越界則返回0，否則返回結果值。

三、代碼實現

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;
    }
}