Reverse Integer-蒲公英云

Reverse Integer

2019独角兽企业重金招聘Python工程师标准>>>

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.

题意：给一个32位整型，要求将其反转，如-123转为-321。注意反转后的前导零和反转后的数字上溢或下溢，如果发生上溢或下溢，则返回0

思路：将整数转为字符数组，从后向前，第一个不为零的数字前的数字全部跳过，逐个将字符添加到一个初始为空的字符串中。最后将通过Integer.parseInt字符串转成int，如果发生上溢和下溢，在转化成int过程中会抛出NumberFormatException异常，如果捕捉到这个异常则返回0，否则返回转换后的数字。

实现：

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;
          }
    }
}