Reverse Integer

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!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

最近做的几题好像都是考数学相关的，所以界限以及判断很重要。

 

如果为最小的负数，应该直接返回0，而不能将其转换为正数之后再转换为负数。

 

C++代码如下：

#include<iostream>
#include<climits>
using namespace std;

class Solution
{
public:
    int reverse(int x)
    {
        //如果x刚好等于INT_MIN,那么不能先将x转换为正数然后转换为负数
        if(x==0||x==INT_MIN)
            return 0;
        int flag=1;
        if(x<0)
        {
            x=-x;
            flag=-1;
        }
        long long i=1;
        long long j=10;
        while(i<=x)
        {
            i*=10;
        }
        i=i/10;
        long long sum=0;
        while(j<=i)
        {
            sum+=(x/i)*(j/10)+(x%j)/(j/10)*i;

            if(flag==1&&sum>INT_MAX)
                return 0;
            else if(flag==-1&&-sum<INT_MIN)
                return 0;
            x=x%i;
            i/=10;
            j*=10;
        }
        //如果是奇数位，中间的数没有加上去，如果位数为奇数位，那么最后i与j相差10倍，否则i与j相差100倍
        if(j==i*10)
            sum+=(x/i)*(j/10);
        if(flag==1&&sum>INT_MAX)
            return 0;
        else if(flag==-1&&-sum<INT_MIN)
            return 0;
        return flag*sum;
    }
};
int main()
{
    Solution s;
    cout<<s.reverse(-2147483648)<<endl;
}

 

方法二：

    int reverse(int x) {
        if(x==0||x==INT_MIN)
            return 0;
        int flag=1;
        if(x<0)
        {
            x=-x;
            flag=-1;
        }
        long long res=0;
        while(x)
        {
            res=res*10+x%10;
            x/=10;
        }
        if((flag==1&&res>INT_MAX)||-res<INT_MIN)
            return 0;
        return flag*res;
    }