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，但是这样要考虑的情况比较多，而且使用了新的空间，效率不高。

可以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;
	}

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

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LeetCode——Reverse Integer

原文：http://blog.csdn.net/laozhaokun/article/details/32944079