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title="JVM核心知识点八股文">JVM核心知识点八股文</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-12-11T07:30:00.000Z" title="发表于 2025-12-11 15:30:00">2025-12-11</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.627Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E5%90%8E%E7%AB%AF%E9%9D%A2%E8%AF%95/">后端面试</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/JVM/">JVM</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%9E%83%E5%9C%BE%E5%9B%9E%E6%94%B6/">垃圾回收</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%86%85%E5%AD%98%E6%A8%A1%E5%9E%8B/">内存模型</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%85%AB%E8%82%A1%E6%96%87/">八股文</a></span></div><div class="content">JVM 核心知识点八股文（高频面试版）本文整理 JVM 面试中最常考察的核心知识点，覆盖内存结构、垃圾回收、类加载、性能调优等高频考点，适合面试复习和知识点梳理。 一、JVM 内存结构（运行时数据区）JDK 8 及以上版本的内存结构移除了永久代（PermGen），替换为元空间（Metaspace），核心分为以下区域： 1. 线程私有区域 程序计数器：记录当前线程执行的字节码行号，唯一不会抛出 OOM 的区域； 虚拟机栈：存储方法调用的栈帧（局部变量表、操作数栈、动态链接、方法出口），栈深度溢出抛 StackOverflowError，内存不足抛 OOM； 本地方法栈：为 Native 方法服务，与虚拟机栈逻辑一致，同样会抛 StackOverflowError/OOM。 2. 线程共享区域 堆：JVM 最大的内存区域，存储对象实例和数组，是垃圾回收（GC）的核心区域； 分代：新生代（Eden + Survivor0 + Survivor1，比例 8:1:1）、老年代； 溢出：对象无法分配内存时抛 OutOfMemoryError: Java heap space。 ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/25/LeetCode1015-%E5%8F%AF%E8%A2%ABK%E6%95%B4%E9%99%A4%E7%9A%84%E6%9C%80%E5%B0%8F%E6%95%B4%E6%95%B0/" title="LeetCode1015.可被K整除的最小整数">LeetCode1015.可被K整除的最小整数</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-25T12:30:00.000Z" title="发表于 2025-11-25 20:30:00">2025-11-25</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.627Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/">哈希表</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E5%AD%A6/">数学</a></span></div><div class="content">1015.可被K整除的最小整数题目描述：给定正整数 k ，你需要找出可以被 k 整除的、仅包含数字 1 的最 小 正整数 n 的长度。 返回 n 的长度。如果不存在这样的 n ，就返回-1。 注意： n 可能不符合 64 位带符号整数。 示例 1: 输入: k = 1 输出: 1 解释: 最小的答案是 n = 1，其长度为 1。 示例 2: 输入: k = 2输出: -1解释: 不存在可被 2 整除的正整数 n 。 示例 3: 输入: k = 3输出: 3解释: 最小的答案是 n = 111，其长度为 3。 提示: 1 <= k <= 10^5 思路： 循环遍历乘10取余k，为了避免溢出对k取余,对k取余不会影响结果 代码：123456789101112class Solution { public int smallestRepunitDivByK(int k) { int res = 0; for (int i = 1; i < 100000; ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/24/LeetCode1018-%E5%8F%AF%E8%A2%AB5%E6%95%B4%E9%99%A4%E7%9A%84%E4%BA%8C%E8%BF%9B%E5%88%B6/" title="LeetCode1018.可被5整除的二进制">LeetCode1018.可被5整除的二进制</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-24T14:56:00.000Z" title="发表于 2025-11-24 22:56:00">2025-11-24</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E4%BD%8D%E8%BF%90%E7%AE%97/">位运算</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E7%BB%84/">数组</a></span></div><div class="content">1018.可被5整除的二进制题目描述：给定一个二进制数组 nums ( 索引从0开始 )。 我们将xi 定义为其二进制表示形式为子数组 nums[0..i] (从最高有效位到最低有效位)。 例如，如果 nums =[1,0,1] ，那么 x0 = 1, x1 = 2, 和 x2 = 5。返回布尔值列表 answer，只有当 xi 可以被 5 整除时，答案 answer[i] 为 true，否则为 false。 示例 1: 输入: nums = [0,1,1] 输出: [true,false,false] 解释: 输入数字为 0, 01, 011；也就是十进制中的 0, 1, 3 。只有第一个数可以被 5 整除，因此 answer[0] 为 true 。 示例 2: 输入: nums = [1,1,1]输出: [false,false,false] 提示: 1 <= nums.length <= 105 nums[i] 仅为 0 或 1 思路： 循环遍历左移一位取余5，为了避免溢出对5取余 代码：1234567891011class Solu...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/23/LeetCode1262-%E5%8F%AF%E8%A2%AB%E4%B8%89%E6%95%B4%E9%99%A4%E7%9A%84%E6%9C%80%E5%A4%A7%E5%92%8C/" title="LeetCode1262.可被三整除的最大和">LeetCode1262.可被三整除的最大和</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-23T14:29:00.000Z" title="发表于 2025-11-23 22:29:00">2025-11-23</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E7%BB%84/">数组</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E8%B4%AA%E5%BF%83/">贪心</a></span></div><div class="content">1262.可被三整除的最大和题目描述：给你一个整数数组 nums，请你找出并返回能被三整除的元素 最大和。 示例 1: 输入: nums = [3,6,5,1,8] 输出: 18 解释: 选出数字 3, 6, 1 和 8，它们的和是 18（可被 3 整除的最大和）。 示例 2: 输入: nums = [4]输出: 0解释： 4 不能被 3 整除，所以无法选出数字，返回 0。 示例 3: 输入: nums = [1,2,3,4,4]输出: 12解释： 选出数字 1, 3, 4 以及 4，它们的和是 12（可被 3 整除的最大和）。 提示: 1 <= nums.length <= 4 * 104 1 <= nums[i] <= 10^4 思路： 虽然A过了，但是我写的贪心太丑陋了，放下灵神的代码吧 代码：123456789101112131415161718192021222324252627282930class Solution { public int maxSumDivThree(int[] nu...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/22/LeetCode65-%E6%9C%89%E6%95%88%E6%95%B0%E5%AD%97/" title="LeetCode65.有效数字">LeetCode65.有效数字</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-22T17:30:00.000Z" title="发表于 2025-11-23 01:30:00">2025-11-23</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/">字符串</a></span></div><div class="content">65.有效数字题目描述：给定一个字符串 s ，返回 s 是否是一个 有效数字。 例如，下面的都是有效数字：2, 0089, -0.1, +3.14, 4., -.9, 2e10, -90E3, 3e+7, +6e-1, 53.5e93, -123.456e789，而接下来的不是：abc, 1a, 1e, e3, 99e2.5, --6, -+3, 95a54e53。 一般的，一个 有效数字 可以用以下的规则之一定义： 一个 整数 后面跟着一个 可选指数。 一个 十进制数 后面跟着一个 可选指数。 一个 整数 定义为一个 可选符号 - 或 + 后面跟着 数字。一个 十进制数 定义为一个 可选符号 - 或 + 后面跟着下述规则： 数字 后跟着一个 小数点 .。 数字 后跟着一个 小数点 . 再跟着 数位。 一个 小数点 . 后跟着 数位。指数 定义为指数符号 e 或 E，后面跟着一个 整数。 数字 定义为一个或多个数位。 示例 1: 输入: s = “0” 输出: true 示例 2: 输入: s = “e”输出: false 示例 3: 输入: s...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/22/LeetCode32-%E6%9C%80%E9%95%BF%E6%9C%89%E6%95%88%E6%8B%AC%E5%8F%B7/" title="LeetCode32.最长有效括号">LeetCode32.最长有效括号</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-22T16:29:00.000Z" title="发表于 2025-11-23 00:29:00">2025-11-23</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/">字符串</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%A0%88/">栈</a></span></div><div class="content">LeetCode32.最长有效括号题目描述：给你一个只包含 ( 和 ) 的字符串，找出最长有效（格式正确且连续）括号 子串 的长度。 左右括号匹配，即每个左括号都有对应的右括号将其闭合的字符串是格式正确的，比如 (()())。 示例 1: 输入: s = “(()” 输出: 2 解释: 最长有效括号子串是 “()” 示例 2: 输入: s = “)()())”输出: 4解释：最长有效括号子串是 “()()” 示例 3: 输入: s = “”输出: 0 提示: 0 <= s.length <= 3 * 104 s[i] 为 ( 或 ) 思路： 栈存下标，每次入栈取栈顶元素看当前待入栈元素是否为),且栈顶元素为(，若满足条件则先弹出元素，然后取ans = Math.max(i-stack(top),ans)需要注意下边界条件，即栈为空边界 代码：1234567891011121314151617181920class Solution { public int longestValidParentheses(Stri...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/22/LeetCode3190-%E4%BD%BF%E6%89%80%E6%9C%89%E5%85%83%E7%B4%A0%E9%83%BD%E5%8F%AF%E4%BB%A5%E8%A2%AB3%E6%95%B4%E9%99%A4%E7%9A%84%E6%9C%80%E5%B0%91%E6%93%8D%E4%BD%9C%E6%95%B0/" title="LeetCode3190.使所有元素都可以被3整除的最少操作数">LeetCode3190.使所有元素都可以被3整除的最少操作数</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-22T15:44:00.000Z" title="发表于 2025-11-22 23:44:00">2025-11-22</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E7%BB%84/">数组</a></span></div><div class="content">3190.使所有元素都可以被3整除的最少操作数题目描述：给你一个整数数组 nums 。一次操作中，你可以将 nums 中的 任意 一个元素增加或者减少 1 。 请你返回将 nums 中所有元素都可以被 3 整除的 最少 操作次数。 示例 1: 输入: nums = [1,2,3,4] 输出: 3 解释: 通过以下 3 个操作，数组中的所有元素都可以被 3 整除： 将 1 减少 1 。 将 2 增加 1 。 将 4 减少 1 。 示例 2: 输入: nums = [3,6,9]输出: 0 提示: 1 <= nums.length <= 50 1 <= nums[i] <= 50 思路： 暴力破解就好，循环代替思考 代码：123456789class Solution { public int minimumOperations(int[] nums) { return Arrays.stream(nums) .map(num -> { ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/21/LeetCode1930-%E9%95%BF%E5%BA%A6%E4%B8%BA3%E7%9A%84%E4%B8%8D%E5%90%8C%E5%9B%9E%E6%96%87%E5%AD%90%E5%BA%8F%E5%88%97/" title="LeetCode1930.长度为3的不同回文子序列">LeetCode1930.长度为3的不同回文子序列</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-21T14:56:00.000Z" title="发表于 2025-11-21 22:56:00">2025-11-21</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/">哈希表</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/">字符串</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%9E%9A%E4%B8%BE/">枚举</a></span></div><div class="content">1930.长度为3的不同回文子序列题目描述：给你一个字符串 s ，返回 s 中 长度为 3 的不同回文子序列 的个数。 即便存在多种方法来构建相同的子序列，但相同的子序列只计数一次。 回文 是正着读和反着读一样的字符串。 子序列 是由原字符串删除其中部分字符（也可以不删除）且不改变剩余字符之间相对顺序形成的一个新字符串。 例如，"ace" 是 "abcde" 的一个子序列。 示例 1: 输入: s = “aabca” 输出: 3 解释: 长度为 3 的 3 个回文子序列分别是： “aba” (“aabca” 的子序列) “aaa” (“aabca” 的子序列) “aca” (“aabca” 的子序列) 示例 2: 输入: s = “adc”输出: 0解释：“adc” 不存在长度为 3 的回文子序列。 示例 3: 输入: s = “bbcbaba”输出: 4解释： 长度为 3 的 4 个回文子序列分别是： “bbb” (“bbcbaba” 的子序列) “bcb” (“bbcbaba” 的子序列) “...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/19/LeetCode757-%E8%AE%BE%E7%BD%AE%E4%BA%A4%E9%9B%86%E5%A4%A7%E5%B0%8F%E8%87%B3%E5%B0%91%E4%B8%BA2/" title="LeetCode757.设置交集大小至少为2">LeetCode757.设置交集大小至少为2</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-19T16:56:00.000Z" title="发表于 2025-11-20 00:56:00">2025-11-20</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E7%BB%84/">数组</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E8%B4%AA%E5%BF%83/">贪心</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%8E%92%E5%BA%8F/">排序</a></span></div><div class="content">757.设置交集大小至少为2题目描述：给你一个二维整数数组 intervals ，其中 intervals[i] = [starti, endi] 表示从 starti 到 endi 的所有整数，包括 starti 和 endi 。 包含集合 是一个名为 nums 的数组，并满足 intervals 中的每个区间都 至少 有 两个 整数在 nums 中。 例如，如果 intervals = [[1,3], [3,7], [8,9]] ，那么 [1,2,4,7,8,9] 和 [2,3,4,8,9] 都符合 包含集合 的定义。 返回包含集合可能的最小大小。 示例 1: 输入: intervals = [[1,3],[3,7],[8,9]] 输出: 5 解释: nums = [2, 3, 4, 8, 9].可以证明不存在元素数量为 4 的包含集合。 示例 2: 输入: intervals = [[1,3],[1,4],[2,5],[3,5]]输出: 3解释： nums = [2, 3, 4].可以证明不存在元素数量为 2 的包含集合。 ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2025/11/19/LeetCode2154-%E5%B0%86%E6%89%BE%E5%88%B0%E7%9A%84%E5%80%BC%E4%B9%98%E4%BB%A52/" title="LeetCode2154. 将找到的值乘以2">LeetCode2154. 将找到的值乘以2</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2025-11-19T09:15:00.000Z" title="发表于 2025-11-19 17:15:00">2025-11-19</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2025-12-11T06:48:10.628Z" title="更新于 2025-12-11 14:48:10">2025-12-11</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/">LeetCode</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/LeetCode/%E7%AE%97%E6%B3%95/">算法</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/algorithm/">算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/">哈希表</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E7%BB%84/">数组</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%8E%92%E5%BA%8F/">排序</a></span></div><div class="content">2154. 将找到的值乘以2题目描述： 给你一个整数数组 nums ，另给你一个整数 original ，这是需要在 nums 中搜索的第一个数字。 接下来，你需要按下述步骤操作： 如果在 nums 中找到 original ，将 original 乘以 2 ，得到新 original（即，令 original = 2 * original）。否则，停止这一过程。 只要能在数组中找到新 original ，就对新 original 继续 重复 这一过程。 返回 original 的 最终 值。 示例 1: 输入: nums = [5,3,6,1,12], original = 3 输出: 24 解释: 3 能在 nums 中找到。3 * 2 = 6 。 6 能在 nums 中找到。6 * 2 = 12 。 12 能在 nums 中找到。12 * 2 = 24 。 24 不能在 nums 中找到。因此，返回 24 。 示例 2: 输入: nums = [2,7,9], original = 4 输...</div></div></div></div><nav id="pagination"><div class="pagination"><span class="page-number current">1</span><a class="page-number" href="/page/2/#content-inner">2</a><span class="space">…</span><a class="page-number" href="/page/4/#content-inner">4</a><a class="extend next" rel="next" href="/page/2/#content-inner"><i class="fas fa-chevron-right fa-fw"></i></a></div></nav></div><div class="aside-content" id="aside-content"><div class="card-widget card-info text-center"><div class="avatar-img"><img src="/img/load.gif" data-lazy-src="/img/avatar.jpg" onerror="this.onerror=null;this.src='/img/friend_404.gif'" alt="avatar"/></div><div class="author-info-name">GoofySatoshi</div><div 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