LeetCode:翻转字符串(7. Reverse Integer)

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本文详细解析了LeetCode题目7：整数反转的解决方案，通过一个高效的算法实现将输入的32位有符号整数进行反转，同时处理了溢出情况，确保在特定的整数范围内正确返回结果。

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LeetCode:7. Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.
Example 1:

Input: 123
Output: 321

Example 2:

put: -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.

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