LeetCode 007 Reverse Integer

题目描述：Reverse Integer

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.

Update (2014-11-10):

Test cases had been added to test the overflow behavior.

Example Questions Candidate Might Ask:

Q: What about negative integers?

A: For input x = –123, you should return –321.

Q: What if the integer’s last digit is 0? For example, x = 10, 100, …

A: Ignore the leading 0 digits of the reversed integer. 10 and 100 are both reversed as 1.

Q: What if the reversed integer overflows? For example, input x = 1000000003.

A: In this case, your function should return 0.

代码如下：

1class Solution {
2public:
3    int reverse(int x) {
4        
5        int ret = 0;
6        
7        while(x != 0){
8            // handle overfolow or underflow
9            if(abs(ret) > 214748364){
10                return 0;
11            }
12            ret = ret * 10 + x % 10;
13            x /= 10;
14        }
15        return ret;
16        
17    }
18};

 Java:

1public int reverse(int x) {
2
3        int ret = 0;
4
5        while (x != 0) {
6
7            // 防止溢出，只要大于MAX/10就算溢出！
8            if (Math.abs(ret) > Integer.MAX_VALUE / 10) {
9                return 0;
10            }
11            ret = ret * 10 + x % 10;
12            x /= 10;
13        }
14
15        return ret;
16    }

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