数字反转（Reverse Integer）

题目：

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?

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).

题意：把要求的数字反转，去掉前置零。

先把数字的符号存到flag中，在将数字从个位开始读取，用temp保存每次读取的数字并进位，再将原数的个位去掉。

遍历一次，正好构造的数就是原数反转，不需要太多额外空间，也没必要转化成字符串。

class Solution {
public:
    int reverse(int x) {
        int temp=0,flag=1;
        if(x<0)flag=-1,x=-x;
        while(x) {
            temp*=10;
            temp+=x%10;
            x/=10;
        }
        return temp*flag;
    }
};