本文主要是介绍面试算法十问2（中英文），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
算法题 1: 数组和字符串
Q: How would you find the first non-repeating character in a string?
问：你如何找到字符串中的第一个不重复字符？
Explanation: Use a hash table to store the count of each character, then iterate through the string to find the first character with a count of one.
解释： 使用哈希表存储每个字符的计数，然后遍历字符串找到计数为一的第一个字符。
function findFirstNonRepeatingChar(string):charCount = {}for char in string:if char in charCount:charCount[char] += 1else:charCount[char] = 1for char in string:if charCount[char] == 1:return charreturn null
算法题 2: 链表
Q: How do you reverse a singly linked list without using extra space?
问：你如何在不使用额外空间的情况下反转一个单链表？
Explanation: Iterate through the list and reverse the links between nodes.
解释： 遍历列表并反转节点之间的链接。
function reverseLinkedList(head):previous = nullcurrent = headwhile current is not null:nextTemp = current.nextcurrent.next = previousprevious = currentcurrent = nextTempreturn previous
算法题 3: 树和图
Q: What is a depth-first search (DFS) and how would you implement it for a graph?
问：什么是深度优先搜索（DFS）？你将如何为一个图实现它？
Explanation: DFS is an algorithm for traversing or searching tree or graph data structures. It starts at the root and explores as far as possible along each branch before backtracking.
解释： DFS是一种用于遍历或搜索树或图数据结构的算法。它从根开始，沿每个分支尽可能深入地探索，然后回溯。
function DFS(node, visited):if node is in visited:returnvisited.add(node)for each neighbor in node.neighbors:DFS(neighbor, visited)
算法题 4: 排序和搜索
Q: Describe how quicksort works and mention its time complexity.
问：描述快速排序是如何工作的，并提及其时间复杂度。
Explanation: Quicksort works by selecting a ‘pivot’ element from the array and partitioning the other elements into two sub-arrays, according to whether they are less than or greater than the pivot. The sub-arrays are then sorted recursively.
解释： 快速排序通过从数组中选择一个“基准”元素，并根据其他元素是小于还是大于基准，将它们划分为两个子数组。然后递归地排序这些子数组。
function quicksort(array, low, high):if low < high:pivotIndex = partition(array, low, high)quicksort(array, low, pivotIndex - 1)quicksort(array, pivotIndex + 1, high)
Time Complexity: Average case is O(n log n), worst case is O(n^2).
时间复杂度： 平均情况是O(n log n)，最坏情况是O(n^2)。
算法题 5: 动态规划
Q: How would you solve the knapsack problem using dynamic programming?
问：你将如何使用动态规划解决背包问题？
Explanation: Create a 2D array to store the maximum value that can be obtained with the given weight. Fill the table using the previous computations.
解释： 创建一个二维数组来存储给定重量可以获得的最大值。使用之前的计算结果填充表格。
function knapsack(values, weights, capacity):n = length(values)dp = array of (n+1) x (capacity+1)for i from 0 to n:for w from 0 to capacity:if i == 0 or w == 0:dp[i][w] = 0elif weights[i-1] <= w:dp[i][w] = max(values[i-1] + dp[i-1][w-weights[i-1]], dp[i-1][w])else:dp[i][w] = dp[i-1][w]return dp[n][capacity]
算法题 6: 数学和统计
Q: How do you compute the square root of a number without using the sqrt function?
问：如何在不使用 sqrt 函数的情况下计算一个数的平方根？
Explanation: Use a numerical method like Newton’s method to approximate the square root.
解释： 使用牛顿法等数值方法来近似计算平方根。
function sqrt(number):if number == 0 or number == 1:return numberthreshold = 0.00001 # Precision thresholdx = numbery = (x + number / x) / 2while abs(x - y) > threshold:x = yy = (x + number / x) / 2return y
算法题 7: 并发编程
Q: Explain how you would implement a thread-safe singleton pattern in Java.
问：解释你将如何在Java中实现一个线程安全的单例模式。
Explanation: Use the initialization-on-demand holder idiom, which is thread-safe without requiring special language constructs.
解释： 使用初始化需求持有者惯用法，它在不需要特殊语言构造的情况下是线程安全的。
public class Singleton {private Singleton() {}private static class LazyHolder {static final Singleton INSTANCE = new Singleton();}public static Singleton getInstance() {return LazyHolder.INSTANCE;}
}
算法题 8: 设计问题
Q: How would you design a system that scales horizontally?
问：你会如何设计一个可以水平扩展的系统？
Explanation: Design the system to work with multiple instances behind a load balancer, use stateless services, and distribute the data across a database cluster.
解释： 设计系统使其能够在负载均衡器后面使用多个实例，使用无状态服务，并在数据库集群中分布数据。
// No specific code, but architectural principles:
- Use load balancers to distribute traffic.
- Implement microservices for scalability.
- Use a distributed database system.
- Employ caching and message queues to handle load.
算法题 9: 实用工具
Q: Write a function to check if a string is a palindrome.
问：编写一个函数检查字符串是否是回文。
Explanation: Compare characters from the beginning and the end of the string moving towards the center.
解释： 比较从字符串开始和结束向中心移动的字符。
function isPalindrome(string):left = 0right = length(string) - 1while left < right:if string[left] != string[right]:return falseleft += 1right -= 1return true
算法题 10: 编码实践
Q: How would you find all permutations of a string?
问：你如何找出一个字符串的所有排列？
Explanation: Use backtracking to swap characters and generate all permutations.
解释： 使用回溯法交换字符并生成所有排列。
function permute(string, l, r):if l == r:print stringelse:for i from l to r:swap(string[l], string[i])permute(string, l+1, r)swap(string[l], string[i]) // backtrack
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