[LeetCode] Reverse Integer 翻转整数

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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

/**
 * Correct but can refactor the code.
 */
class Solution {
public:
    int reverse(int x) {
        long long res = 0;
        bool isPositive = true;
        if (x < 0) {
            isPositive = false;
            x *= -1;
        }
        while (x > 0) {
            res = res * 10 + x % 10;
            x /= 10;
        }
        if (res > INT_MAX) return 0;
        if (isPositive) return res;
        else return -res;
    }
};

提交通过后，OJ给出了官方解答，一看比自己的写的更精简一些，它没有特意处理正负号，仔细一想，果然正负号不影响计算，而且没有用long long型数据，感觉写的更好一些，那么就贴出来吧：

class Solution {
public:
    int reverse(int x) {
        int res = 0;
        while (x != 0) {
            if (abs(res) > INT_MAX / 10) return 0;
            res = res * 10 + x % 10;
            x /= 10;
        }
        return res;
    }
};

在贴出答案的同时，OJ还提了一个问题 To check for overflow/underflow, we could check if ret > 214748364 or ret < –214748364 before multiplying by 10. On the other hand, we do not need to check if ret == 214748364, why? （214748364 即为 INT_MAX / 10）

为什么不用check是否等于214748364呢，因为输入的x也是一个整型数，所以x的范围也应该在 -2147483648～2147483647 之间，那么x的第一位只能是1或者2，翻转之后res的最后一位只能是1或2，所以res只能是 2147483641 或 2147483642 都在int的范围内。但是它们对应的x为 1463847412 和 2463847412，后者超出了数值范围。所以当过程中res等于 214748364 时， 输入的x只能为 1463847412， 翻转后的结果为 2147483641，都在正确的范围内，所以不用check。