reverse int

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:
Input: 123
Output: 321


Example 2:
Input: -123
Output: -321


Example 3:
Input: 120
Output: 21


Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

class Solution {
public:
    int reverse(int x) {
        // edge case
        // 对于integer的操作，要考虑其overflow的情况。
        // max, numeric_limits<int>::max() .. min() .. 
        if (x == numeric_limits<int>::max() || x == numeric_limits<int>::min()) {
            return 0;
        }
        
        if (x >= 0) {
            return reverseUtil(x);
        }
        else {
            return 0 - reverseUtil(abs(x));
        }
    }
    
    int reverseUtil(int x)
    {
        long reverse = 0;
        while(x != 0) {
            reverse = reverse*10 + x%10;
            x /= 10;
        }
        
        // edge case
        // 对于integer的操作，要考虑其overflow的情况。
        // max, numeric_limits<int>::max() .. min() .. 
        if (reverse > numeric_limits<int>::max()) {
            return 0;
        }
        
        return static_cast<int>(reverse);
    }
};

错误总结：

1. 记得考虑numeric的边界和overflow的情况

2. reverse数字的巧妙方法。

        long reverse = 0;
        while(x != 0) {
            reverse = reverse*10 + x%10;
            x /= 10;
        }