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: [-2147483648,2147483647]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

解題思路：

該題目思路簡單，要注意的是int型別的整數範圍是[-2147483648,2147483647]，因此當-2147483648和2147483647取反後將不再是int型別，因此解決該題目的思路是可以先將傳進來的數值賦值給long long型變數，然後對long long型變數進行取反，如果取反後的數值大於INT_MAX，則應返回0。C++程式碼如下：

class Solution {
public:
    int reverse(int x) 
    {
        long long y = x;
        bool flag = true;
        if
 
(y<0)
        {
            flag = false;
            y = -y;
        }
        long long res = 0;
        while(y>0)
        {
            res = res * 10+y % 10;
            y = y / 10;
        }
        if(res>INT_MAX)
        {
            return 0;
        }
        if(flag)
        {
            return
 
 res;
        }
        else
        {
            return -res;
        }
    }
};

執行結果：