【LeetCode】Reverse Integer

最新推荐文章于 2021-01-12 00:40:21 发布

ice4026

最新推荐文章于 2021-01-12 00:40:21 发布

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分类专栏：算法学习【LeetCode】

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本文介绍了一种简单的整数翻转算法，并提供了C++代码实现。该算法能够将输入的整数（如123）翻转为新的整数（如321）。文章还讨论了翻转过程中可能遇到的问题，例如溢出处理和末尾0的处理。

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题目描述：

Reverse digits of an integer.

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

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).

题目没啥好说的。关于后面的几个问题，我想还是应该在不同的实际环境中分别处理吧。可以返回exception，可以返回0，也可以用个long来作为返回值。

代码如下：

class Solution {
public:
	int reverse(int x) {
		int res(0);
		while (x){
			res = res * 10 + x % 10;
			x = x / 10;
		}
		return res;
	}
};