LeetCode——Reverse Integer-优快云博客

本文介绍了一个LeetCode上的经典题目“逆序整数”的解决方案，通过逐步迭代的方式实现整数的逆序，同时讨论了处理边界情况的方法。

LeetCode——Reverse Integer

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).

数字逆转。

可转成字符数组。再处理。可是这样要考虑的情况比較多，并且使用了新的空间，效率不高。

能够通过求模运算获得依次低位，再把原乘以10加上新取得的数。

	public static int reverse(int x) {
		int ret = 0;
		while (x != 0) {
			ret = ret * 10 + x % 10;
			x /= 10;
		}
		return ret;
	}

比方：输入-192，计算步骤例如以下，经过了三轮循环：

ret x
---------------
0 -192
-2 -192
-2 -19
---------------
-2 -19
-29 -19
-29 -1
---------------
-29 -1
-291 -1
-291 0
---------------
-291

posted on 2017-07-21 15:31 mthoutai 阅读( ...) 评论( ...) 编辑收藏

转载于:https://www.cnblogs.com/mthoutai/p/7217901.html