5 - 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).

solution： 此题简单，但需考虑两点，一是x的正负，引入ifNeg这个bool类型值来表示，然后reverse循环只是不停取当前数的最后一位然后累加至目标数*10的结果，在循环中应当添加判断是否越界的语句（正数越界后小于零），此处我将其放置在循环外，只图效率。

class Solution {
public:

    int reverse(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        bool ifNeg = false;
        if(x < 0)
        {
            ifNeg = true;
            x = abs(x);
        }
        
        int result = 0;
        while(x > 0)
        {
            int num = x % 10;
            result = result * 10 + num;
            x = x/10;
        }
        if(result < 0)
            return -1;
        
        if(ifNeg)
            result *= -1;
        
        return result;
    }
};