7. Reverse Integer
7. Reverse Integer
题目
Given a signed 32-bit integer x
, return x
with its digits reversed. If reversing x
causes the value to go outside the signed 32-bit integer range [-2^31, 2^31 - 1]
, then return 0
.
Assume the environment does not allow you to store 64-bit integers (signed or unsigned).
Example 1:
Input: x = 123
Output: 321
Example 2:
Input: x = -123
Output: -321
Example 3:
Input: x = 120
Output: 21
Constraints:
-2^31 <= x <= 2^31 - 1
题目大意
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。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。
解题思路
- 这一题是简单题,要求反转 10 进制数。类似的题目有 第 190 题。
- 这一题只需要注意一点,反转以后的数字要求在
[−2^31, 2^31 − 1]
范围内,超过这个范围的数字都要输出0
。
代码
/**
* @param {number} x
* @return {number}
*/
var reverse = function (x) {
const isNegative = x > 0 ? 1 : -1;
let temp = 0;
x *= isNegative;
while (x > 0) {
temp = temp * 10 + (x % 10);
x = Math.floor(x / 10);
if (temp < -(2 ** 31) || temp > 2 ** 31 - 1) return 0;
}
return temp * isNegative;
};
相关题目
相关题目
- [8. 字符串转换整数 (atoi)](./0008.md)
- [190. 颠倒二进制位](https://leetcode.com/problems/reverse-bits)
- [2119. 反转两次的数字](https://leetcode.com/problems/a-number-after-a-double-reversal)
- [2442. 反转之后不同整数的数目](https://leetcode.com/problems/count-number-of-distinct-integers-after-reverse-operations)