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?

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.

題意：給一個32位整型，要求將其反轉，如-123轉爲-321。注意反轉後的前導零和反轉後的數字上溢或下溢，若是發生上溢或下溢，則返回0

思路：將整數轉爲字符數組，從後向前，第一個不爲零的數字前的數字所有跳過，逐個將字符添加到一個初始爲空的字符串中。最後將經過Integer.parseInt字符串轉成int，若是發生上溢和下溢，在轉化成int過程當中會拋出NumberFormatException異常，若是捕捉到這個異常則返回0，不然返回轉換後的數字。

實現：

public class Solution {
    public int reverse(int x) {
        int r=0;
          char[]t=(x+"").toCharArray();
          boolean flag=false;
          String result="";
          int s=1;
          for(int i=t.length-1;i>=0;i--){
               if(!flag&&t[i]=='0')
                    continue;
               else
                    flag=true;
               if(i==0&&t[i]=='-'){
                    s*=-1;
                    break;
               }
               result+=t[i];
          }
          try{
               r=Integer.parseInt(result) ;
               return r*s;
          }catch(NumberFormatException e){
               return 0;
          }
    }
}