[Leetcode] Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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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).

[思路及要点]

1. 很多题解开辟isNegative标志，在最后将结果根据标志进行处理是否要*1，其实这里是不需要的，负号对于计算的结果没有影响。

2. 对于溢出的处理，虽然测试样例没有包含。使用long long存储结果，一旦越界及时跳出。

[代码]

class Solution {public:    int reverse(int x) {        long long res = 0;        while (x) {            res = res * 10 + x % 10;            if (res > INT_MAX || res < INT_MIN) return -1;            x /= 10;        }                return res;    }};

从今天开始，伴随着自己的第二轮跳槽准备，我将会把之前的所有代码重新优化并且提供解题思路，欢迎大家参与讨论！

11/02/2014