Reverse digits of an integer.

Example1:x = 123, return 321

Example2:x = -123, return -321

Spoilers:

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).

## Analysis

This is a simple problem. But spoilers are very important. The test cases on LeetCode do not test the overflow case. This should be specified if we are doing this problem.

The range of integer in Java is -2^{32} to 2^{32} – 1. If we keep number adding to Integer.MAX_VALUE, it will be a negative number until it turns to zero. This can be used to test the overflow. But in another problem, this can not be used. I will post it if I reach that problem.

## Code

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public class Solution { public int reverse(int x) { boolean sign = false; if (x < 0) { sign = true; x = -x; } int ret = 0; while (x > 0) { int tmp = x % 10; ret = ret * 10 + tmp; x /= 10; } if (sign) return -ret; else return ret; } } |

The code is very simple. Just use mod operation to get the last digit of original number and multiply the result with 10 and add the digit. The sign of the number should be considered.

## Complexity

The complexity of this algorithm is O(log n).