## Monday, December 31, 2012

### [LeetCode] Reverse Integer 解题报告

Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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).
» Solve this problem

[解题思路]

[Code]
``````1:    int reverse(int x) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      int lastDigit = 0;
5:      int result = 0;
6:      bool isNeg = x>0? false:true;
7:      x = abs(x);
8:      while(x>0)
9:      {
10:        lastDigit = x%10;
11:        result = result*10 + lastDigit;
12:        x = x/10;
13:      }
14:      if(result<0) return -1;
15:      if(isNeg)
16:        result *=-1;
17:      return result;
18:    }
``````

Version 2, Refactor code, 3/5/2012
Rethink for a while. And actually, we don't need to use special logic to handle '+' and '-', because the sign can be kept during calculating. Only 8 lines can resolve this problem.
``````1:       int reverse(int x) {
2:            int newN =0, left =0;
3:            while(x != 0)
4:            {
5:                 left = x%10;
6:                 newN = newN*10 + left;
7:                 x = x/10;
8:            }
9:            return newN;
10:       }
``````