算法第二次作业

整数反转与溢出处理

最新推荐文章于 2024-01-19 11:27:25 发布

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本文介绍了一种整数反转的方法，并详细讨论了在32位整数环境下如何处理反转过程中可能出现的溢出问题。文章提供了一段C++代码实现，通过检查新结果与旧结果之间的关系来避免溢出导致的错误。

题目：Reverse Integer
描述：
Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

click to show spoilers.
Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

题解：
将数字进行反转，这个so easy，不断的取余，乘十就行。
需要注意的是判断溢出的操作，我的解决办法是溢出会导致乱码生成，在循环中判断。不过还有一种判断方法
(j > INT_MAX || j < INT_MIN) ? 0: (int)j;
代码：
class Solution { public: int reverse(int x) { int result=0; while(x!=0) { int tail=x%10; int newresult=result*10+tail; if((newresult-tail)/10!=result) {return 0;} result=newresult; x=x/10; } return result; } };