leetcode 7 Reverse Integer

7. Reverse Integer

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Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

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将int类型的数字翻转过来。按最简单的模10处理。

注意：倒转过来的时候，可能会导致溢出问题。

int类型的范围是[-65536, 65535]

例如： 一个数为10009，倒过来为90001，就会导致溢出。

​ 同理一个数为-10009，同样也会导致溢出。

所以需要问清楚，当溢出时返回结果是什么。

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class Solution {
  public int reverse(int x) {
      long result = 0;
      while (x != 0){
          int tail = x % 10;
          result = result*10 + tail;
          if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE){
              return 0;
          }
          x = x/10;
      }
      return (int) result;
  }
}