算法題 1: 數(shù)組和字符串

Q: How would you find the first non-repeating character in a string?
問：你如何找到字符串中的第一個不重復(fù)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.
解釋： 使用哈希表存儲每個字符的計數(shù)，然后遍歷字符串找到計數(shù)為一的第一個字符。

function findFirstNonRepeatingChar(string):
    charCount = {}
    for char in string:
        if char in charCount:
            charCount[char] += 1
        else:
            charCount[char] = 1
    
    for char in string:
        if charCount[char] == 1:
            return char
    return null

算法題 2: 鏈表

Q: How do you reverse a singly linked list without using extra space?
問：你如何在不使用額外空間的情況下反轉(zhuǎn)一個單鏈表？

Explanation: Iterate through the list and reverse the links between nodes.
解釋： 遍歷列表并反轉(zhuǎn)節(jié)點之間的鏈接。

function reverseLinkedList(head):
    previous = null
    current = head
    while current is not null:
        nextTemp = current.next
        current.next = previous
        previous = current
        current = nextTemp
    return previous

算法題 3: 樹和圖

Q: What is a depth-first search (DFS) and how would you implement it for a graph?
問：什么是深度優(yōu)先搜索（DFS）？你將如何為一個圖實現(xiàn)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.
解釋： DFS是一種用于遍歷或搜索樹或圖數(shù)據(jù)結(jié)構(gòu)的算法。它從根開始，沿每個分支盡可能深入地探索，然后回溯。

function DFS(node, visited):
    if node is in visited:
        return
    visited.add(node)
    for each neighbor in node.neighbors:
        DFS(neighbor, visited)

算法題 4: 排序和搜索

Q: Describe how quicksort works and mention its time complexity.
問：描述快速排序是如何工作的，并提及其時間復(fù)雜度。

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.
解釋： 快速排序通過從數(shù)組中選擇一個“基準(zhǔn)”元素，并根據(jù)其他元素是小于還是大于基準(zhǔn)，將它們劃分為兩個子數(shù)組。然后遞歸地排序這些子數(shù)組。

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).
時間復(fù)雜度： 平均情況是O(n log n)，最壞情況是O(n^2)。

算法題 5: 動態(tài)規(guī)劃

Q: How would you solve the knapsack problem using dynamic programming?
問：你將如何使用動態(tài)規(guī)劃解決背包問題？

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.
解釋： 創(chuàng)建一個二維數(shù)組來存儲給定重量可以獲得的最大值。使用之前的計算結(jié)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] = 0
            elif 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]

算法題 6: 數(shù)學(xué)和統(tǒng)計

Q: How do you compute the square root of a number without using the sqrt function?
問：如何在不使用 sqrt 函數(shù)的情況下計算一個數(shù)的平方根？

Explanation: Use a numerical method like Newton’s method to approximate the square root.
解釋： 使用牛頓法等數(shù)值方法來近似計算平方根。

function sqrt(number):
    if number == 0 or number == 1:
        return number
    threshold = 0.00001  # Precision threshold
    x = number
    y = (x + number / x) / 2
    while abs(x - y) > threshold:
        x = y
        y = (x + number / x) / 2
    return y

算法題 7: 并發(fā)編程

Q: Explain how you would implement a thread-safe singleton pattern in Java.
問：解釋你將如何在Java中實現(xiàn)一個線程安全的單例模式。

Explanation: Use the initialization-on-demand holder idiom, which is thread-safe without requiring special language constructs.
解釋： 使用初始化需求持有者慣用法，它在不需要特殊語言構(gòu)造的情況下是線程安全的。

public class Singleton {
    private Singleton() {}

    private static class LazyHolder {
        static final Singleton INSTANCE = new Singleton();
    }

    public static Singleton getInstance() {
        return LazyHolder.INSTANCE;
    }
}

算法題 8: 設(shè)計問題

Q: How would you design a system that scales horizontally?
問：你會如何設(shè)計一個可以水平擴(kuò)展的系統(tǒng)？

Explanation: Design the system to work with multiple instances behind a load balancer, use stateless services, and distribute the data across a database cluster.
解釋： 設(shè)計系統(tǒng)使其能夠在負(fù)載均衡器后面使用多個實例，使用無狀態(tài)服務(wù)，并在數(shù)據(jù)庫集群中分布數(shù)據(jù)。

// 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.

算法題 9: 實用工具

Q: Write a function to check if a string is a palindrome.
問：編寫一個函數(shù)檢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.
解釋： 比較從字符串開始和結(jié)束向中心移動的字符。

function isPalindrome(string):
    left = 0
    right = length(string) - 1
    while left < right:
        if string[left] != string[right]:
            return false
        left += 1
        right -= 1
    return true

算法題 10: 編碼實踐

Q: How would you find all permutations of a string?
問：你如何找出一個字符串的所有排列？

Explanation: Use backtracking to swap characters and generate all permutations.
解釋： 使用回溯法交換字符并生成所有排列。文章來源地址http://www.zghlxwxcb.cn/news/detail-857683.html

function permute(string, l, r):
    if l == r:
        print string
    else:
        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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