摘要：October 25,2025 Today is Saturday,and it's windy.I went to the company to work as usual,but this is considered overtime.Because there are two days to 阅读全文

摘要：Welcome to my channel Bookish English 2! Have you ever dreamed of speaking English… but felt too shy, too slow, or just not good enough? Well, I have 阅读全文

摘要：Welcome to my channel, Bookish English 2. Do you study English, learn grammar, memorize words, but when it's time to speak, nothing comes out? Your mi 阅读全文

摘要：At my level, an influencer with this many followers, if I do a live stream PK battle, the money I'd make would be easy. Rain or shine, I'll be waiting 阅读全文

摘要：when I was in gomi particularly I felt very very close to a great spiritual [Music] power if we think of the Bible it would be in which we live and mo 阅读全文

摘要：Welcome to Bookish English 2. Tell me honestly, how many times have you said, "I'll start learning English tomorrow." But tomorrow never comes, right? 阅读全文

摘要：Welcome to my channel, Bookish English 2. Today I will show you something real. Last week I met a student who spoke English with only 50 sentences. Sh 阅读全文

摘要：Welcome back to my channel, Bookish English. Today's video is all about the most powerful English books that every beginner must read before 2026. The 阅读全文

摘要：Hello my friends and welcome back to Bookish English 2, the place where you live English, not just study it. Today's lesson is special. You will not r 阅读全文

摘要：给你一个非负整数数组 nums ，你最初位于数组的 第一个下标 。数组中的每个元素代表你在该位置可以跳跃的最大长度。 判断你是否能够到达最后一个下标，如果可以，返回 true ；否则，返回 false 。 class Solution { public boolean canJump(int[] n 阅读全文

摘要：给定一个数组 prices ，它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。 你只能选择 某一天 买入这只股票，并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。 返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0 阅读全文

摘要：中位数是有序整数列表中的中间值。如果列表的大小是偶数，则没有中间值，中位数是两个中间值的平均值。 例如 arr = [2,3,4] 的中位数是 3 。 例如 arr = [2,3] 的中位数是 (2 + 3) / 2 = 2.5 。 实现 MedianFinder 类: MedianFinder() 阅读全文

摘要：LeetCode热题100-73题、前K个高频元素 给你一个整数数组 nums 和一个整数 k ，请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。 class Solution { public int[] topKFrequent(int[] nums, int k) { // 阅读全文