leetcode学习笔记07

问题思考方法1

问题

reverse-integer

Given a 32-bit signed integer, reverse digits of an integer.

Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Example 1:

Input: x = 123 Output: 321

Example 2:

Input: x = -123 Output: -321

Example 3:

Input: x = 120 Output: 21

Example 4:

Input: x = 0 Output: 0

Constraints:

$-2^{31}<=x<=2^{31}-1$

思考

这道题的难度是简单.确实不难,按位反转就好了.但是怎么处理边际值是一个问题, 简单的按位反转在计算过程中可能会碰到越界的情况, 怎么判断越界是个问题.

方法1

一个比较讨巧的方法是, 在计算过程中我用double来储存中间值, 最后看结果是否超过了int的范围, 没超过就转型为int返回结果,超过了就返回0;

时间复杂度$O(1)$. 空间复杂度$O(1)$.

class Solution { public int reverse(int x) { double res = 0; while(x != 0){ res *= 10; res += x % 10; x = x / 10; } if(res < Integer.MIN_VALUE || res > Integer.MAX_VALUE) return 0; return (int)res; } }