在本文中，我们将探讨Meta（前Facebook）面试中的两道真题，并提供详细的解析和思路。这些题目不仅考察了数据结构和算法的基本功，还涉及了概率计算。希望通过这篇文章，能帮助你更好地准备面试。
In this blog, we will explore two genuine Meta (formerly Facebook) interview questions and provide detailed analysis and approaches. These questions test not only fundamental knowledge of data structures and algorithms but also involve probability calculations. Through this article, we hope to help you better prepare for your interviews.
题目1：二叉搜索树中寻找最接近目标值的节点
Question 1: Closest Value in a Binary Search Tree
题目英文原文：
Meta Q1: Given a binary search tree, and an integer target, return the node whose value is closest to the target.
题目描述：
给定一个二叉搜索树（BST）和一个整数目标值，返回树中节点值最接近目标值的节点。
分析：
- 二叉搜索树特性：BST中的左子节点值总是小于根节点值，右子节点值总是大于根节点值。这一特性可以帮助我们有效地缩小搜索范围。
- 遍历策略：我们可以使用递归或迭代的方法来遍历树。从根节点开始，根据当前节点值与目标值的比较结果，决定向左子树还是右子树继续搜索。
- 更新最接近值：在遍历过程中，持续更新与目标值最接近的节点值，直到遍历完成。
Analysis:
- BST Properties: In a BST, the value of the left child node is always less than the root node's value, and the value of the right child node is always greater than the root node's value. This property helps efficiently narrow down the search range.
- Traversal Strategy: We can use either recursive or iterative methods to traverse the tree. Starting from the root, depending on the comparison result of the current node value and the target value, decide to move to the left or right subtree.
- Updating Closest Value: During traversal, keep updating the value of the node that is closest to the target value until the traversal is complete.
题目2：基于人口生成城市名称的概率计算
Question 2: City Generation Probability Based on Population
题目英文原文：
Given a list of city names and their corresponding populations, write a program to output a city name subject to the following constraint: the probability of the program to output a city's name is based on its population divided by the sum of all cities' population. Assume the program will be repeatedly called many times.
For example:
- NY: 7M
- SF: 5M
- LA: 8M
The probability to generate NY is 7/20, SF is 5/20, and LA is 8/20.
题目描述：
给定一组城市名称及其对应的人口数量，编写一个程序，以城市人口占总人口的比例为概率，输出一个城市名称。假设程序会被反复调用多次。
示例：
- 输入：
- NY: 7M
- SF: 5M
- LA: 8M
- 输出概率：
- 生成NY的概率是 7/20
- 生成SF的概率是 5/20
- 生成LA的概率是 8/20
分析：
- 概率计算：每个城市生成的概率等于其人口数除以总人口数。我们需要预先计算这些概率，并存储它们以便快速查询。
- 前缀和数组：为了在常数时间内实现加权随机选择，可以使用前缀和数组。前缀和数组的每个元素表示从第一个城市到当前城市的累计概率。
- 随机选择：生成一个在0到总人口数之间的随机数，通过二分查找在前缀和数组中找到对应的城市。
Analysis:
- Probability Calculation: The probability of generating each city is equal to its population divided by the total population. We need to pre-calculate these probabilities and store them for quick access.
- Prefix Sum Array: To achieve constant time complexity for weighted random selection, we can use a prefix sum array. Each element in the prefix sum array represents the cumulative probability from the first city to the current city.
- Random Selection: Generate a random number between 0 and the total population, and use binary search on the prefix sum array to find the corresponding city.
示例讲解
示例1：BST最近目标值
- 假设给定的BST为：markdown复制代码
4 / \ 2 5 / \ 1 3目标值为3.714。 - 遍历顺序：
- 从根节点4开始，3.714 < 4，向左子树移动。
- 到达节点2，3.714 > 2，向右子树移动。
- 到达节点3，3.714 > 3，但没有右子树，遍历结束。
- 结果：节点值3最接近3.714。
示例2：城市生成概率
- 给定城市及人口：
- NY: 7M
- SF: 5M
- LA: 8M
- 总人口：20M
- 概率计算：
- NY: 7/20
- SF: 5/20
- LA: 8/20
- 前缀和数组：[7, 12, 20]
- 随机数生成及查找：
- 生成一个0到20之间的随机数，通过二分查找找到对应城市。
总结
这两道题目分别考察了树结构的遍历和概率计算的实现。通过充分理解二叉搜索树的特性和前缀和数组的应用，可以高效解决这些问题。希望本文的详细解析能够帮助你在面试中取得好成绩。我们提供面试辅助，代面试服务，助您进入梦想大厂，欢迎随时咨询我。
Conclusion
These two questions test the traversal of tree structures and the implementation of probability calculations. By thoroughly understanding the properties of binary search trees and the application of prefix sum arrays, these problems can be efficiently solved. We hope this detailed analysis helps you perform well in your interviews.
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