(C++練習) 7. Reverse Integer

題目 :

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

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.

大意 :

逆轉一個intege數字, 包含正負號, note 有提示要小心超出邊界值。

 1 class Solution {
 2 public:
 3     int reverse(int x) {
 4 
 5         try
 6         {
 7             bool isNeg = x > 0 ? true : false;
 8             int lastDigit;
 9             int reverse = abs(x);
10             int res = 0;
11             while (reverse)
12             {
13                 if (res > 214748364 || 
14                     res < -214748364 ||
15                     (res == 214748364 && x % 10 > 7) ||
16                     (res == -214748364 && x % 10 < -8))
17                 {
18                     return 0;
19                 }
20                 lastDigit = reverse % 10;
21                 res = res * 10 + lastDigit;
22                 reverse = reverse / 10;
23             }
24 
25             if (isNeg)
26                 return res;
27             else
28                 return res * -1;
29         }
30         catch (exception e)
31         {
32             return 0;
33         }
34 
35     }
36 };

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