7. Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321Example2: 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.

需要考虑到溢出的问题以及100的问题，如-2147483648，这个数本身就溢出了，需要先转成long，然后再做处理，我的方法比较笨，就是先把数转成正数，然后在最后计算结果的时候将符号再乘上去。

public class Solution {    public int reverse(int x) {        List<Long> list = new ArrayList<>();        int flag = 1;        if(x < 0) {            flag = -1;        }        long absx = Math.abs((long) x);        long quotient = absx/10, remainder = absx%10;        while(quotient != 0) {            list.add(remainder);            remainder = quotient %10;            quotient = quotient /10;        }        list.add(remainder);        long result = 0;        for(int k = 0, j = list.size()-1; k<list.size(); k++,j--) {            result += list.get(k) * Math.pow(10, j);        }        return result > Integer.MAX_VALUE ? 0 : (int)(result * flag);    }}

看了别人提交的答案，差距啊。。。mark一下学习，感觉思考的深度不够，实现太简单，导致写起来就很麻烦。尤其是中间的部分，空间复杂度大很多。

public int reverse(int x) {        long rev= 0;        while( x != 0){            rev= rev*10 + x % 10;            x= x/10;            if( rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE)                return 0;        }        return (int) rev;    }