We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? # for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “#”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? # to your account
Difficulty: 中等
Related Topics: 数学
给你一个 32 位的有符号整数 x ,返回将 x 中的数字部分反转后的结果。
x
如果反转后整数超过 32 位的有符号整数的范围 [−231, 231 − 1] ,就返回 0。
假设环境不允许存储 64 位整数(有符号或无符号)。
示例 1:
输入:x = 123 输出:321
示例 2:
输入:x = -123 输出:-321
示例 3:
输入:x = 120 输出:21
示例 4:
输入:x = 0 输出:0
提示:
Language: JavaScript
/** * @param {number} x * @return {number} */ var reverse = function(x) { let result = 0 while(x) { result = result * 10 + x % 10 x = x / 10 | 0 } return (result | 0) === result ? result : 0 };
The text was updated successfully, but these errors were encountered:
No branches or pull requests
7. 整数反转
Description
Difficulty: 中等
Related Topics: 数学
给你一个 32 位的有符号整数
x
,返回将x
中的数字部分反转后的结果。如果反转后整数超过 32 位的有符号整数的范围 [−231, 231 − 1] ,就返回 0。
假设环境不允许存储 64 位整数(有符号或无符号)。
示例 1:
示例 2:
示例 3:
示例 4:
提示:
Solution
Language: JavaScript
The text was updated successfully, but these errors were encountered: