Leetcode007 Reverse Integer

Problem:

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 store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Intuition:

easy tag,本身要说的内容不多。本题唯一要注意的也就是溢出部分的处理。在java中，integer.MAXVALUE比integer.MINVALUE的绝对值要小1，这个地方如果忽略掉的话就很容易出错。

下列代码中的7是Integer.MAXVALUE除以10的余数。8是Integer.MINVALUE的绝对值除以10的余数。

Solution:

public int reverse(int x) {
      int rev = 0;
     while (x != 0) {
         int pop = x % 10;
         x /= 10;
         if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
         if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
         rev = rev * 10 + pop;
     }
     return rev;
 }