题目描述：给定一个 32 位有符号整数，将整数中的数字进行反转。

方法：弹出和推入数字 & 溢出前进行检查

思路

我们可以一次构建反转整数的一位数字。在这样做的时候，我们可以预先检查向原整数附加另一位数字是否会导致溢出。

方案一

#include<iostream>
using namespace std;
//常规方案
int reverse(int number)
{
	int receive = 0;
	while (number != 0)
	{
		int pop = number % 10;
		number /= 10;
		if (receive > INT_MAX / 10 || (receive == INT_MAX / 10 && pop > 7))
		{//防止最大整数溢出
			return 0;
		}
		if (receive < INT_MIN / 10 || (receive == INT_MIN / 10 && pop < -8))
		{//防止最小负整数溢出
			return 0;
		}
		receive = receive * 10 + pop;
	}
	return receive;
}

int main()
{
	//测试数据
	int number1 = 1463847412;
	int number2 = reverse(number1);
	cout << number2 << endl;
	int number3 = -1463847412;
	int number4 = reverse(number3);
	cout << number4 << endl;
	//溢出示例
	cout << INT_MAX + INT_MAX << endl;
	cout << INT_MIN + INT_MIN << endl;
	cout << INT_MAX + 1 << endl;
	cout << INT_MIN - 1 << endl;
	cout << INT_MAX + INT_MIN << endl;
	return 0;
}

结果展示： 

方案二：

#include<iostream>
using namespace std;
//非主流方案(效率高一点)
int reverse(int number)
{
	int sum = 0;
	while (number != 0)
	{
		int num = sum * 10;
		if (num / 10 != sum)
		{//在这儿判断是否溢出
			return 0;
		}
		sum = num;
		sum += number % 10;
		number = number / 10;
	}
	return sum;
}

int main()
{
	//测试数据
	int number1 = 1463847412;
	int number2 = reverse(number1);
	cout << number2 << endl;
	int number3 = -1463847412;
	int number4 = reverse(number3);
	cout << number4 << endl;
	//溢出示例
	cout << INT_MAX + INT_MAX << endl;
	cout << INT_MIN + INT_MIN << endl;
	cout << INT_MAX + 1 << endl;
	cout << INT_MIN - 1 << endl;
	cout << INT_MAX + INT_MIN << endl;
	return 0;
}

结果展示：