Reverse Integer:将给定数字倒序返回

Reverse digits of an integer.

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

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

输入样例包括：9646324351这种超过32位的数字

思路一：就当成普通字符串，颠倒顺序处理即可，注意当位数不符int类型时返回0即可。

public int reverse(int x) {
        try{
            String s = x+"";
            StringBuilder sb = new StringBuilder();
            int index = 0;
        if(s.charAt(0)==('-'))
        {
            sb.append("-");
            index++;
        }
        for(int j = s.length()-1;j>=index;j--){
            sb.append(s.charAt(j)+"");
        }
            return Integer.parseInt(sb.toString());
        }catch(Exception e){
            return 0;
        }
       
    }

思路二：当成正常的数值进行运算，诸位取出（取余）在存入新数，并判断是否超过位数

public int reverse(int x)
{
    int result = 0;

    while (x != 0)
    {
        int tail = x % 10;
        int newResult = result * 10 + tail;
        if ((newResult - tail) / 10 != result)
        { return 0; }
        result = newResult;
        x = x / 10;
    }

    return result;
}

思路二参考：https://leetcode.com/problems/reverse-integer/discuss/