LeetCode-Reverse Integer

Reverse Integer

##题目

####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.

##解题思路
该题是将给定的整数进行逆序输出，可以通过对10进行取余而得到其个位，十位，百位等。
但是需要注意以下三个问题：

1.对于负数的逆序，只需要对其正数进行翻转，然后添加符号即可。
2.类似于100这样的数，最后的结果应该是1，而不是001.
3.逆转之后可能会产生比原数更大的数，从而造成整数的溢出。

##算法代码
代码采用JAVA实现：

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public class Solution {
    public int reverse(int x) {
        if(x==0) return 0;
        if(x<0) return -reverse(-x);
        //每次取模相加，考虑到末尾为0的情况
        int reverseNum=0;
        int modnum=0;
        while(x>0)
        {
            modnum=x%10;
            //考虑翻转会溢出的情况
            if((reverseNum>Integer.MAX_VALUE/10)||((reverseNum==Integer.MAX_VALUE/10)&&(modnum>Integer.MAX_VALUE%10)))
                return 0;
            reverseNum=reverseNum*10+modnum;
            x=x/10;
        }
        return reverseNum;
    }
}

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