【Reverse Integer】cpp

题目：

Reverse digits of an integer.

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

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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.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

代码：

class Solution {
public:
    int reverse(int x) {
            int v = x;
            queue<int> que;
            while ( true )
            {
                int digit = v % 10;
                que.push(digit);
                v = v / 10;
                if ( abs(v)<10 ) break;
            }
            if ( v!=0) que.push(v);
            int ret = 0;    
            while ( !que.empty() )
            {
                // cout << que.front() << endl;
                if ( ret > INT_MAX/10 || ret < INT_MIN/10) return 0; // overflow
                ret = ret * 10 + que.front();
                que.pop();
            }
            return ret;
    }
};

tips：

主要考虑出现overflow怎么处理，上述代码第一次写用了queue的结构，先进先出，方便调试各种case。

AC后对上述的代码进行了改进，改进结果如下，也可以AC。

class Solution {
public:
    int reverse(int x) {
            int ret = 0;
            while (x)
            {
                if ( ret > INT_MAX/10 || ret < INT_MIN/10 ) return 0;
                ret = ret*10 + x%10;
                x = x /10;
            }
            return ret;
    }
};

可以看到代码精炼了很多。另外判断是否越界不要直接比较 value>INT_MAX或者value<INT_MIN一般都是比较最大值除以10；因为value本身就越界了，再跟最大值比较也没有意义了。这是一个放缩的技巧。

=============================================

第二次过这道题，把corner case考虑完全了，就AC了。

class Solution {
public:
    int reverse(int x) {
            vector<int> digits;
            int sign = x>0 ? 1 : -1;
            int remain = 1;
            while ( remain>0 )
            {
                digits.push_back(fabs(x%10));
                x = x/10;
                remain = fabs(x);
            }
            int ret = 0;
            for ( int i=0; i<digits.size(); ++i )
            {
                if ( INT_MAX/10<ret || (INT_MAX/10==ret && INT_MAX%10<digits[i]) )
                {
                    return 0;
                }
                ret = ret*10 + digits[i];
            }
            return ret*sign;
    }
};