7 Reverse Integer 整数反转
Description:
Given a 32-bit signed integer, reverse digits of an integer.
Example:
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
题目描述:
给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。
示例:
示例 1:
输入: 123
输出: 321
示例 2:
输入: -123
输出: -321
示例 3:
输入: 120
输出: 21
注意:
假设我们的环境只能存储得下 32 位的有符号整数，则其数值范围为 [−2^31, 2^31 − 1]。请根据这个假设，如果反转后整数溢出那么就返回 0。
思路:
整型范围: -2147483648～2147483647
- 可以用字符串反转, 注意判断是否溢出即可
- 用取余和整数除法(地板除)
时间复杂度O(lgn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int reverse(int x)
{
int result = 0;
while (x != 0)
{
int last = x % 10;
x /= 10;
if (result > INT_MAX / 10 or (result == INT_MAX / 10 and last > 7)) return 0;
if (result < INT_MIN / 10 or (result == INT_MIN / 10 and last < -8)) return 0;
result = result * 10 + last;
}
return result;
}
};
Java:
class Solution {
public int reverse(int x) {
int result = 0;
while (x != 0) {
int last = x % 10;
x /= 10;
if (result > Integer.MAX_VALUE / 10 || (result == Integer.MAX_VALUE / 10 && last > 7)) return 0;
if (result < Integer.MIN_VALUE / 10 || (result == Integer.MIN_VALUE / 10 && last == -9)) return 0;
result = 10 * result + last;
}
return result;
}
}
Python:
class Solution:
def reverse(self, x: int) -> int:
if not x:
return 0
str_x = str(x)
x = ''
if (str_x[0] == '-'):
x += '-'
x += str_x[::-1].lstrip('0').rstrip('-')
else:
x += str_x[::-1].lstrip('0')
x = int(x)
if x > 2 ** 31 - 1 or x < -2 ** 31:
return 0
return x
