1、题目：

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).

2、解答：

public class Solution {public int reverse(int x) {int out = 0;//保存当前结果int newout = 0;//保存最新结果while(x!=0){newout = out*10 + x%10;//目前最新计算的结果=上次结果*10+新的尾数if((newout-x%10)/10!=out){//逆运算判断是否溢出return 0;}x = x/10;//降位out = newout;}return out;}
}

3、思路：

最简单的做法是：

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

但存在溢出问题，int型有自己的取值范围。

2的解法加了一步逆运算，

if((newout-x%10)/10!=out)，即用当前的最新结果newout-刚才加上的尾数x%10，若不等于out*10，则说明存在溢出问题。

溢出是指，newout = out*10 + x%10的值超过最大位数时，系统会将超过最大位数的高位舍弃，从而导致反向验算时出现(newout-x%10)/10!=out。