7 Reverse Integer

Method: 简单模拟，注意几种情况处理，前导的0的处理，负数的处理。

class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        if x == 0:
            return 0
        if x > 0:
            if x > 2**31-1:
                return 0
            x = str(x)
            ans = ""
            mark = 1
            for i in range(1,len(x)+1):
                if mark and x[len(x)-i] =='0':
                    continue
                else:
                    mark = 0
                ans += x[len(x)-i]
            if int(ans) > 2**31-1:
                return 0
            else:
                return int(ans)
        if x < 0:
            if x < -(2**31):
                return 0
            x = str(x)
            x = x[1:]
            mark = 1
            ans = "-"
            for i in range(1,len(x)+1):
                if mark and x[len(x)-i] == '0':
                    continue
                else:
                    mark = 0
                ans = ans + x[len(x)-i]
            if int(ans) < -(2**31):
                return 0
            return int(ans)