Question ：


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).


Anwser 1 ：
 

class Solution {
public:
    int reverse(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ret = 0;
        
        int div = 1;
        bool flag = false;
        if(x < 0) {         // flag a negative
            flag = true;
            x = -x;
        }
        
        queue<int> Q;
        while(x > 0){       
            int mod = x % 10;   // from low digit to high
            Q.push(mod);
            
            x /= 10;
            div *= 10;
        }
        
        while(!Q.empty()){
            int mod = Q.front();    // pop low digit
            Q.pop();
            
            div /= 10;
            ret = ret + mod * div;
        }
        
        return flag ? -ret : ret;
    }
};


Anwser 2 ：
 

class Solution {
public:
    int reverse(int x) {
        int res = 0;
        
        bool flag = x < 0 ? true : false;
        x = flag ? -x : x;
        
         while (x > 0) {       // don't care positive or negetive
             res = res * 10 + x % 10;   // get lowest digit then multi 10
             x /= 10;
         }
         
         return flag ? -res : res;
    }
};


more simple ：

class Solution {
public:
    int reverse(int x) {
        int res = 0;
         
         while (x != 0) {       // don't care positive or negetive
             res = res * 10 + x % 10;   // get lowest digit then multi 10
             x /= 10;
         }
         
         return res;
    }
};