leetcode7

public class Solution {
    public int Reverse(int x) {
        int fuhao = 1;
            if (x < 0)
            {
                fuhao = -1;
            }

            try
            {
                x = Math.Abs(x);
            }
            catch (Exception e)
            {
                return 0;
            }

            int i = 0;

            var list = new List<long>();

            do
            {
                long num = x % Convert.ToInt64(Math.Pow(10, i + 1)) / Convert.ToInt64(Math.Pow(10, i));
                list.Add(num);
                i++;
            }
            while (x / Convert.ToInt64(Math.Pow(10, i)) != 0);

            var length = list.Count;

            long result = 0;

            for (int j = 0; j < length; j++)
            {
                try
                {
                    result += list[j] * Convert.ToInt64(Math.Pow(10, length - j - 1));
                }
                catch (Exception e)
                {
                    result = 0;
                    break;
                }
            }

            result *= fuhao;

            int res = 0;

            int.TryParse(result.ToString(), out res);

            //Console.WriteLine(res);
            return res;
    }
}

https://leetcode.com/problems/reverse-integer/#/description

补充一个python的实现：

class Solution:
    def reverse(self, x: int) -> int:
        if x == 0:
            return 0
        operation = 1
        if x < 0:
            operation = -1
        sx = str(x)
        if sx[0] == '-':
            sx = sx[1:]
        n = len(sx)
        result = []
        for i in range(n-1,-1,-1):
            result.append(sx[i])
        r = int(''.join(result))
        r *= operation
        if r >= 2 ** 31 or r < (-2) ** 31:
            r = 0
        return r