[LintCode] Reverse Integer 翻轉整數

 

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上的原題，請參見我之前的博客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 = 0;
        while (n != 0) {
            res = 10 * res + n % 10;
            n /= 10;
        }
        return (res < INT_MIN || res > INT_MAX) ? 0 : res;
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param n the integer to be reversed
     * @return the reversed integer
     */
    int reverseInteger(int n) {
        int res = 0;
        while (n != 0) {
            int t = res * 10 + n % 10;
            if (t / 10 != res) return 0;
            res = t;
            n /= 10;
        }
        return res;
    }
};