LeetCode 7. Reverse Integer

原题

Reverse digits of an integer.

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

click to show spoilers.

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.

易错点分析

1. 如果输入的值结尾是0，那么逆置后你应该输出什么？应该是去除0后的值。
2. 判断是否超过32位最大整形数应该放到最后面，因为有可能出现输入值没有超界，逆置后超界的情况。

思路分析

1. 列表逆置法：将数字转化为列表，再将列表逆置，再转化为数字。
2. 相除取余法：基本的数学取逆置数方法，效率较高

代码实现

# 列表逆置
# Your runtime beats 25.33 % of python submissions.
class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        # return sys.maxint
        if x >= 0:
            ret = reversed(list(str(x)))
            ret = int("".join(ret))
        else:
            x = abs(x)
            ret = reversed(list(str(x)))
            ret = -int("".join(ret))
        return ret if -2147483648 <= ret <= 2147483647 else 0

# 相除取余
# Your runtime beats 76.84 % of python submissions.
class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        # return sys.maxint
        if x >= 0:
            ret = self.solve(x)
        else:
            ret = -self.reverse(-x)
        return ret if -2147483648 <= ret <= 2147483647 else 0
        
    def solve(self, num):
        ret = 0
        while num > 0:
            ret *= 10
            ret += num % 10
            num /= 10
        return ret