Reverse Integer

最新推荐文章于 2021-05-21 05:05:16 发布

原创最新推荐文章于 2021-05-21 05:05:16 发布 · 435 阅读

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本文详细探讨了反转整数的基本算法，并着重分析了处理整数末尾为0的情况以及整数反转可能引起的溢出问题。通过实例演示，展示了如何在不同边界条件下正确实现整数反转功能。

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

click to show spoilers.

Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
1、If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
2、Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

3、Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

题意比较清楚：

反转整数

一般直接转为正数便于处理，返回值的时候做判断即可。

其中的细节是越界（第二条），我的处理是直接输出极值

#include<iostream>
#include<cmath>
#include<climits>
using namespace::std;
class Solution {
public:
    int reverse(int x)
    {
        //abs(INT_MIN)>INT_MAX;
        if(x == INT_MIN)
            return INT_MIN;
        int num = abs(x);
        int res = 0;
        while(num != 0)
        {
            //提早一位处理越界
            if(res>(INT_MAX-num%10)/10)
            return x>0?INT_MAX:INT_MIN;
            //反转的处理
            res = res*10 + num%10;
            num = num/10;
        }
        return x>0?res:(-res);
    }
};
int main()
{
    int x=-123;
    Solution so;
    cout<<so.reverse(x);
}