Given an integer and rotates the number of this integer position

[Question:]

Reverse digits of an integer. EXAMPLE1:X = 123, return 321

example2:x = -123, return-321

Solving: Given an integer, the number of this integer is rotated position.

[Analysis:] Here are some good the questions to ask before coding. Bonus points for/If you have already thought through this!

If the digit is 0, what should the output to be? ie, cases such as 10, 100.

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Did you notice the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should do you handle such cases?

Throw an exception? Good, but what if throwing a exception is isn't an option? You are would then have to re-design the function (ie, add a extra parameter). Note Point 1: If the last digit of an integer is 0, such as: 10100, then the inversion will turn into 101. NOTE 2: If you enter a number of 1000000003, then the reversal will overflow, you need to pay attention.

[Solution:]

* 
 * Assume the input 123: 
 * 123-> 3   
 *-> + 2   
 * 1-> + 1 *   
 * Public
class Solution {  
    pub LIC int reverse (int x) {  
        int = 0;  
        while (x!= 0) {Results  
            = result * + x% 10;//each time on the basis of the original results to 10 times times larger, plus the remainder  
            x = X/10;  
        Eturn result;  
    }  
     
    public static void Main (string[] args) {  
        Scanner Scanner = new Scanner (new Bufferedinputstream (system.in));  
        while (Scanner.hasnext ()) {  
            int num = Scanner.nextint ();  
            System.out.println (New Solution (). reverse (num));}}