7. Reverse Integer--数字反转输出

Reverse digits of an integer.

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

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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?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Note:

The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

这道题目的意思是：输入一个整数，输出它反转之后的数字，如果反转后的数字溢出，则返回0；思路如下：用取余的方式取出输入整数的每一位，然后将取出的数字用newResult（即第n-1次循环取出的数）保存，然后将newResult*10加上tail（第n次循环取出的数），正常情况下，(newResult-tail)/10=result，如果result*10+tail溢出，即大于Integer.MAX_VALUE（2147483647）或小于Integer.MIN_VALUE（-2147483648）时，newResult只能取result*10+tail的低32位，因此这时(newResult-tail)/10!=result，则返回0。

代码如下：

int result = 0;
while (x != 0)

4000
{

int tail = x % 10;
int newResult = result * 10 + tail;
if ((newResult-tail)/10!=result)
{ return 0; }

//保存中间结果
result = newResult;
x = x / 10;
}

return result;