LeetCode 题解工作台
两整数相加
给你两个整数 num1 和 num2 ,返回这两个整数的和。 示例 1: 输入: num1 = 12, num2 = 5 输出: 17 解释: num1 是 12,num2 是 5 ,它们的和是 12 + 5 = 17 ,因此返回 17 。 示例 2: 输入: num1 = -10, num2 = …
1
题型
7
代码语言
3
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简单 · 数学·driven
答案摘要
我们可以直接使用加法运算符 `+` 来计算两个整数的和。 时间复杂度 ,空间复杂度 。
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题目描述
num1 和 num2,返回这两个整数的和。
示例 1:
输入:num1 = 12, num2 = 5 输出:17 解释:num1 是 12,num2 是 5 ,它们的和是 12 + 5 = 17 ,因此返回 17 。
示例 2:
输入:num1 = -10, num2 = 4 输出:-6 解释:num1 + num2 = -6 ,因此返回 -6 。
提示:
-100 <= num1, num2 <= 100
解题思路
方法一:使用加法运算符
我们可以直接使用加法运算符 + 来计算两个整数的和。
时间复杂度 ,空间复杂度 。
class Solution:
def sum(self, num1: int, num2: int) -> int:
return num1 + num2
复杂度分析
| 指标 | 值 |
|---|---|
| 时间 | Depends on the final approach |
| 空间 | Depends on the final approach |
面试官常问的追问
外企场景- question_mark
The candidate should demonstrate an understanding of basic integer operations.
- question_mark
The candidate may be asked about edge cases and how integer overflow is handled.
- question_mark
The candidate should be able to explain the efficiency of their approach.
常见陷阱
外企场景- error
Forgetting to handle edge cases such as extreme negative or positive values.
- error
Overcomplicating the solution by introducing unnecessary logic for the addition operation.
- error
Not taking advantage of the simple nature of the problem, leading to overly complex solutions.
进阶变体
外企场景- arrow_right_alt
What if the input numbers are very large? The principle still holds, though overflow might need consideration in some languages.
- arrow_right_alt
How would you approach this problem if you had to add more than two integers?
- arrow_right_alt
How might this problem change if the range of integers was significantly larger?