Reverse Integer_LeetCode

Description:

Reverse digits of an integer.

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

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?

 

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

 

Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

 

解题思路：

用一个范围比32位int大出很多的long long 变量来存储倒转后的数字。

利用一个while循环来对原数字不断取模得到最后一位数字。

result = result*10+x%10;

为本算法的核心代码。

同时，要对倒转得到的数字检查是否溢出，可以利用C/C++中的limits.h头文件中的INT_MAX和INT_MIN直接进行比较。

 

 

 代码：

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