[Leetcode] Reverse Integer

题目：

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).

思路：对原数从后往前一位一位的取，对结果不断移位。对于符号的问题，整个操作并没有改变符号的操作，因此最终的正负性会保留；对于溢出的问题，将结果设置为long long型，最终判断即可。

class Solution {public:    int reverse(int x) {        long long result = 0;        while (x != 0) {            result = result * 10 + x % 10;            x /= 10;        }        if (result > INT_MAX || result < INT_MIN) return 0;        else return (int)result;    }};

总结：复杂度为O(n).