leetcode-top-interview-questions-easyZa**ny 在2024-11-30 14:13:07 访问0 Bytes 《初级算法》是一本关于算法的入门书籍,它包含了许多经典的算法和问题。在这本书中,作者通过一些简单的代码示例来讲解这些算法和问题。 例如,书中有一个关于排序的问题,要求读者实现一个冒泡排序算法。冒泡排序是一种简单的排序...
welcome to my blog LeetCode Top Interview Questions 9. Palindrome Number (Java版; Easy) 题目描述 Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. Example 1: Input: 121 Output: true Example 2: Input: -121 Output: ...
LeetCode Top Interview Questions 69. Sqrt(x) (Java版; Easy) 题目描述 Implement int sqrt(int x). Compute and return the square root of x, where x is guaranteed to be a non-negative integer. Since the return type is an integer, the decimal digits are truncated and only the integer part...
【Leetcode】【Easy】Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! If the intege...
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Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
For regular desktop internet users, you can get also this Leetcode Night Mode version on your favorite web browser. That is with the Night Mode option in the Turn Off the Lights Browser extension. Follow the steps below on how to enable this free solution in your web b...
* https://leetcode.cn/leetbook/read/top-interview-questions-easy/xnto1s/ * https://leetcode.cn/problems/first-bad-version/solution/di-yi-ge-cuo-wu-de-ban-ben-by-leetcode-s-pf8h/ * </pre> * NB. * * @author skyler_11@163.com ...
But wasn't Amazon the company that expected people to solve Leetcode 1192. Critical Connections in a Network on an OA? And then in another OA expected people to find the Articulation Points in a graph too?Like seriously if you ask those questions on a OA, where a person can't get any...