7. Reverse Integer【easy】

解题思路：

挨个遍历，不断把末位数赋给新的值即可。

考虑边界条件：

当结果溢出时返回0，所以为了不让中间值溢出，采用 long 类型来保存结果。

实现代码：

// 7. Reverse Integer
int reverse(int x) {
  long result = 0, longX = abs((long)x);
  while (longX > 0) {
    result = result * 10 + longX % 10;
    longX /= 10;
  }
  result = (x > 0) ? result : -result;
  if (result > INT32_MAX || result < INT32_MIN) {
    return 0;
  } else {
    return (int)result;
  }
}

问题描述：

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: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.