leetcode: Reverse Integer

最新推荐文章于 2021-01-09 00:40:58 发布

原创最新推荐文章于 2021-01-09 00:40:58 发布 · 767 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#stringstream #C++ #leetcode

C/C++ 同时被 2 个专栏收录

33 篇文章

订阅专栏

Algorithm

29 篇文章

订阅专栏

本文探讨了如何实现整数反转并处理特殊情况，如最后一位为0和整数溢出的问题。通过使用字符串流简化操作，同时讨论了边界情况的处理策略。

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?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

思路

很简单，把所有位的数字都算出来就行了，恩，不少人应该都是这么想的。不过，我用字符串流来解决了这个问题，方便快捷无污染

。不过，貌似运行时间要长那么一点

。关于上面提到的，最后一位是0的处理方法，貌似这两种方法都不要特意处理。关于溢出，我想在返回值是int型数值的情况下，貌似是无法解决的。事实上，在所有的测试案例中貌似也没有可以导致溢出的测试案例。我想，他也是提醒我们注意极端的情况以及边界情况。

class Solution {
public:
    int reverse(int x) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		bool isNegative=false;
        if(x<0) isNegative=true;
		stringstream stream;
		stream<<x;
		string str;
		stream>>str;
		stream.str("");
		stream.clear();
		for(int i=str.length()-1;i>=0;i--)
		{
			stream<<str[i];
		}
		int res;
		stream>>res;
		stream.str("");
		stream.clear();
		if(isNegative) res=-res;
		return res;
    }
};