[LintCode] Reverse Integer 翻转整数

Reverse digits of an integer. Returns 0 when the reversed integer overflows (signed 32-bit integer).

Have you met this question in a real interview?

Example

Given x = 123, return 321

Given x = -123, return -321

LeetCode上的原题，请参见我之前的博客Reverse Integer。

解法一：

class Solution {
public:
    /**
     * @param n the integer to be reversed
     * @return the reversed integer
     */
    int reverseInteger(int n) {
        long long res = ;
        while (n != ) {
            res =  * res + n % ;
            n /= ;
        }
        return (res < INT_MIN || res > INT_MAX) ?  : res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param n the integer to be reversed
     * @return the reversed integer
     */
    int reverseInteger(int n) {
        int res = ;
        while (n != ) {
            int t = res *  + n % ;
            if (t /  != res) return ;
            res = t;
            n /= ;
        }
        return res;
    }
};