[Interview] Morgan Stanley 2026 Intern Apprentice Program
Categories Interview
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面经
开场
面试官告诉了我大致的流程:大摩上海的业务基本上就是面向中国的,所以大部分同事也是中国人,但是在周会的时候需要用英语做presentation。所以面试中自我介绍需要用英文,但是后面的问题可以夹杂着中文。我提到对后面技术方面中文更熟悉,面试官同意。
八股
基本上是结合着项目问的。项目问的不深但是会扣项目中的点来问八股,总之被拷打死了…
- llm原理
- 什么是自注意力
- 用过哪些llm,对不同的llm感受如何
- 用grafana做了什么
- 短连接和长连接
- 什么是op
- 什么是哈希函数
- 什么是哈希表
- 什么是红黑树
- 什么是数据库事务隔离机制
- override overwrite 区别
- 手写sql语句
- 手写redis命令
- redis原理
- go中有哪些处理并发的数据结构
- class变量和object变量的场景
- 对并发的了解
算法
整体都是easy难度,第一题见过但是还是想了七八分钟…
直接在纸上手撕
- 一杯3l 一杯5l 怎么凑出4l
365. Water and Jug Problem - 二叉搜索树 如何插入节点
700. Search in a Binary Search Tree 搜索换成插入罢了 - 如果树是线性的会怎么样
- rand7实现rand20
470. Implement Rand10() Using Rand7() - m x n的网格 从1,1走到m,n有多少路径
62. Unique Paths - 上一题 如果有障碍物
63. Unique Paths II
英语
- Short introduction(开场)
- How do you think about a team(结束)
反问
Many financial institutions are seen as conservative in tech stack selection. However, I noticed that Morgan Stanley invests billions of dollars in technology annually. 那么实习生怎么感受到这一点呢?面试官回答:
- 摩根目前会采用新的技术,但是肯定不会像互联网公司那样采用最新的技术栈。
- manager 会给实习生分配特定的项目,也可以和同事交流。
- 可以在公司内使用内网部署的或者经过审批的模型。开发都在内网,不可以在公网进行。不过可以回家 gemini 一下(
总结
面试官人还是非常非常nice的,但是我的八股能力还是太差了,面试前面的表现非常不好。同时我的英语口语也是一坨,自我介绍大概只有一分钟,最后的英语问题回答的也磕磕巴巴。总结为寄
最后面试官认为我才大二,机会还有很多。整体感觉我还是挺聪明的(,不过还要多重视一下基础。
面试官会给我留下面评,之后如果manager要我就会电话联系,否则就不一定会联系了。
这次还是花太多时间在算法上了(4天),八股其实也都不难,应该好好准备的。同时在邀面之前就应该准备起来(导致最后准备的时间只有5天),不然我今天早上想到要面试都极其痛苦。最后还是要对项目里面的技术要很熟悉才可以(特别是面向对象的内容一定要很熟悉)。英语能力对外企来说还是挺重要的,我这个口语能力感觉还是太不合格了。
八股算法准备
算法复习
- 704. 二分查找
- 59. 螺旋矩阵II
- 44. 开发商购买土地
- 203. 移除链表元素
- 707. 设计链表
- 206. 反转链表
- 142. 环形链表II
- 242. 有效的字母异位词
- 349. 两个数组的交集
- 1. 两数之和 这题从没想到,返回的数组下标一定是最后面的
- 454. 四数相加II 如果是n数相加,时间复杂度是O(n / 2 上取整)
- 54. 替换数字(第八期模拟笔试)
- 151. 翻转字符串里的单词 想了老半天
- 28. 实现 strStr()
- 459.重复的子字符串
- 214. Shortest Palindrome
- 128. Longest Consecutive Sequence
- 15. 三数之和
- 18. 四数之和 去重!!!
- 232. 用栈实现队列
- 225. 用队列实现栈
- 239. 滑动窗口最大值
- 347. 前 K 个高频元素
- 94. Binary Tree Inorder Traversal
- 226. Invert Binary Tree 中序、前后序、层序、递归 四种方法都可以
- 101. 对称二叉树 递归和迭代
- 111. Minimum Depth of Binary Tree 最小深度要在叶子结点处理
- 222. Count Complete Tree Nodes 递归利用完全二叉树性质
- 110. Balanced Binary Tree 高度后序 深度前序
- 513. Find Bottom Left Tree Value 前序遍历 如果交换left和right呢?
- 112. Path Sum 迭代 pair
- 106. Construct Binary Tree from Inorder and Postorder Traversal 用index
- 654. Maximum Binary Tree index
- 617. Merge Two Binary Trees 类似于对称二叉树
- 700. Search in a Binary Search Tree 迭代
- 98. Validate Binary Search Tree 数组 递归 中序迭代 记住中序遍历
- 501. Find Mode in Binary Search Tree 中序 众数处理
- 236. Lowest Common Ancestor of a Binary Tree 回溯 后序遍历
- 235. Lowest Common Ancestor of a Binary Search Tree 第一个区间内就是 注意本身就是祖先的情况
- 701. Insert into a Binary Search Tree 递归 迭代
- 450. Delete Node in a BST 递归 五种情况
- 669. Trim a Binary Search Tree
- 108. Convert Sorted Array to Binary Search Tree 用index构造 构造类题目都有套路 递归
- 538. Convert BST to Greater Tree 反中序遍历
- 77. Combinations 组合 剪枝
- 40. Combination Sum II 去重 used和candidate i-1 i
- 39. Combination Sum 注意参数是i还是i+1还是start
- 216. Combination Sum III 好多边界…
- 131. Palindrome Partitioning
- 470. Implement Rand10() Using Rand7() (randX - 1) * Y + randY = randXY
- 78. Subsets 其实就是终止条件改改
start == size不是错误 - 46. Permutations used数组
nums.size() == p.size() - 491. Non-decreasing Subsequences 不能排序 不能用原来的去重 用set
- 63. Unique Paths II 注意初始化条件
- 343. Integer Break
- 416. Partition Equal Subset Sum 记住01背包公式 边界条件
- 494. Target Sum 边界条件
dp[j] += dp[j - nums[i]] - 200. Number of Islands dfs bfs
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};
void bfs(vector<vector<char>>& grid, int i, int j, vector<vector<bool>>& visited) {
queue<pair<int, int>> q;
q.push({i, j});
visited[i][j] = true;
while (!q.empty()) {
pair<int , int> p = q.front(); q.pop();
auto curi = p.first, curj = p.second;
for (int k = 0; k < 4; k++) {
auto nexti = curi + dir[k][0], nextj = curj + dir[k][1];
if (nexti < 0 || nextj < 0 || nexti >= grid.size() || nextj >= grid[0].size() || grid[nexti][nextj] == '0' || visited[nexti][nextj]) continue;
visited[nexti][nextj] = true;
q.push({nexti, nextj});
}
}
}
void dfs(vector<vector<char>>& grid, int i, int j, vector<vector<bool>> &visited) {
for (int k = 0; k < 4; k++) {
auto nexti = i + dir[k][0], nextj = j + dir[k][1];
if (nexti < 0 || nexti >= grid.size() || nextj < 0 || nextj >= grid[0].size() || visited[nexti][nextj] || grid[nexti][nextj] != '1') {
continue;
}
visited[nexti][nextj] = true;
dfs(grid, nexti, nextj, visited);
}
}
- 994. Rotting Oranges bfs
- 207. Course Schedule 邻接表 入度
日常问题
Why Morgan?
a. Morgan Stanley is a world-leading institution. It’s not just a bank; it’s a global leader that defines the standards of the financial industry.
b. morgan Shanghai has Global standards, interns also have chance to contact to cutting-edge architecture in company
c. luckily I was recommended for this program by my instructor at ECNU. I’ve always focused on Morgan Stanley’s Apprentice əˈprɛntɪs Program, and this recommendation gave me the perfect opportunity to prove my skills.
d. some of my seniors from Boyuan IT Club previously interned at Morgan Stanley. Their positive feedback about the supportive culture and technical excellence here really inspired me to apply.Short Introduction
Introduction (10s)
“Good afternoon, I am Rui Chen, a sophomore majoring in Software Engineering at ECNU. My current GPA is 3.85, ranking in the top 5% of my class.”Experience & Projects (30s)
“I have a strong background in backend development. I’ve led the development of several projects, including ‘Meowpick’, a course review mini-program with over 10,000 users, and an AI-based psychological assessment platform using Large Language Models.”Leadership (20s)
“Beyond coding, I am the President of the Boyuan IT Club. I’ve organized several major events, including a campus-wide Hackathon and practical workshops on LLM applications.”Closing (10s)
“I am passionate about building high-quality systems and very excited about the opportunity to join Morgan Stanley’s Apprentice Program. Thank you!”I have a question regarding technology strategy. Many financial institutions are seen as conservative in tech stack selection. However, I noticed that Morgan Stanley invests billions of dollars in technology annually. How does the firm balance between financial stability and adopting cutting-edge technologies?
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