leetcode

题目：Reverse Integer

Reverse digits of an integer.

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

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?

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).

个人思路：

1、将数字转换成字符串，如果是是负数，则将负号放在字符串末尾，然后字符串反转，再将字符串转换成数字即可

2、上面的思路可以AC，但是没有考虑溢出的问题，由于这个问题并不在test case中，可以作为额外的思考题

代码：

 1 #include <string>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 class Solution
 7 {
 8 public:
 9     int reverse(int x)
10     {
11         char x_chars[64];
12         int x_value = abs(x);
13 
14         if (x < 0)
15         {
16             sprintf(x_chars, "%d-", x_value);
17         }
18         else
19         {
20             sprintf(x_chars, "%d", x_value);
21         }
22         string x_string(x_chars);
23         string x_reverse(x_string.rbegin(), x_string.rend());
24 
25         return atoi(x_reverse.c_str());
26     }
27 };
28 
29 int main()
30 {
31     Solution s;
32     int test = s.reverse(-123490);
33 
34     cout << test << endl;
35 
36     system("pause");
37 
38     return 0;
39 }

View Code

上网看了几篇文章，都是通过除以10和模10的方法来取得尾数，有一遍文章有对溢出问题的思考，链接：http://blog.csdn.net/littlestream9527/article/details/17059939