Java面试突击：分享十份常见面试题及代码案例，附资源下载链接 |21xrx.com

Java面试突击：分享十份常见面试题及代码案例，附资源下载链接

2023-06-17 00:19:26 深夜i -- --

Java面试代码案例资源下载

如果你正在备战Java相关岗位面试，不妨来看看这篇分享。以下是十份常见Java面试题及其代码案例，涵盖了面试中的基础知识、算法和细节方面。希望对大家有所帮助。

1. 请实现一个单例模式。


public class Singleton {
  private Singleton() {}
  private static Singleton instance = null;
  public static synchronized Singleton getInstance() {
    if (instance == null) {
      instance = new Singleton();
    }
    return instance;
  }
}

2. 请实现一个链表的翻转。


public class ReverseList {
  static class Node {
    int data;
    Node next;
    public Node(int data)
      this.data = data;
      this.next = null;
    
  }
  public static Node reverse(Node head) {
    Node pre = null;
    Node curr = head;
    while (curr != null)
      Node next = curr.next;
      curr.next = pre;
      pre = curr;
      curr = next;
    
    return pre;
  }
}

3. 请实现一个二叉树的前序遍历。


public class PreorderTraversal {
  static class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    public TreeNode(int val)
      this.val = val;
      this.left = null;
      this.right = null;
    
  }
  public static void preorderTraversal(TreeNode root) {
    if (root != null) {
      System.out.print(root.val + " ");
      preorderTraversal(root.left);
      preorderTraversal(root.right);
    }
  }
}

4. 请实现一个算法，判断一个字符串是否是回文串。


public class IsPalindrome {
  public static boolean isPalindrome(String s) {
    if (s == null)
      return false;
    
    int left = 0;
    int right = s.length() - 1;
    while (left < right) {
      char c1 = s.charAt(left);
      char c2 = s.charAt(right);
      if (!Character.isLetterOrDigit(c1)) {
        left++;
      } else if (!Character.isLetterOrDigit(c2))
        right--;
       else {
        if (Character.toLowerCase(c1) != Character.toLowerCase(c2))
          return false;
        
        left++;
        right--;
      }
    }
    return true;
  }
}

5. 请实现一个冒泡排序算法。


public class BubbleSort {
  public static void bubbleSort(int[] arr) {
    if (arr == null || arr.length <= 1)
      return;
    
    for (int i = 0; i < arr.length - 1; i++) {
      for (int j = 0; j < arr.length - 1 - i; j++) {
        if (arr[j] > arr[j + 1]) {
          int temp = arr[j];
          arr[j] = arr[j + 1];
          arr[j + 1] = temp;
        }
      }
    }
  }
}

6. 请实现一个HashMap。


public class MyHashMap
  {
 
  private static final int DEFAULT_CAPACTIY = 16;
  private static final float DEFAULT_LOAD_FACTOR = 0.75f;
  private Node
  [] array;
 
  private int size;
  @SuppressWarnings("unchecked")
  public MyHashMap() {
    array = (Node
  []) new Node[DEFAULT_CAPACTIY];
 
    size = 0;
  }
  public void put(K key, V value) {
    int index = hash(key);
    Node
  head = array[index];
 
    Node
  node = head;
 
    while (node != null) {
      if (equalsKey(node.key, key)) {
        node.value = value;
        return;
      }
      node = node.next;
    }
    Node
  newNode = new Node<>(key, value);
 
    newNode.next = head;
    array[index] = newNode;
    size++;
    if (needRehashing()) {
      rehashing();
    }
  }
  public V get(K key) {
    int index = hash(key);
    Node
  node = array[index];
 
    while (node != null) {
      if (equalsKey(node.key, key)) {
        return node.value;
      }
      node = node.next;
    }
    return null;
  }
  private boolean equalsKey(K k1, K k2) {
    return k1 == null ? k2 == null : k1.equals(k2);
  }
  private int hash(K key) {
    if (key == null) {
      return 0;
    }
    int code = key.hashCode();
    return code & (array.length - 1);
  }
  private boolean needRehashing() {
    float loadFactor = (float) size / array.length;
    return loadFactor >= DEFAULT_LOAD_FACTOR;
  }
  @SuppressWarnings("unchecked")
  private void rehashing() {
    Node
  [] oldArray = array;
 
    array = (Node
  []) new Node[oldArray.length * 2];
 
    size = 0;
    for (Node
  head : oldArray) {
 
      Node
  node = head;
 
      while (node != null) {
        put(node.key, node.value);
        node = node.next;
      }
    }
  }
  static class Node
  {
 
    K key;
    V value;
    Node
  next;
 
    public Node(K key, V value) {
      this.key = key;
      this.value = value;
      this.next = null;
    }
  }
}

7. 请实现一个十进制转二进制的算法。


public class DecimalToBinary {
  public static String decimalToBinary(int n) {
    StringBuilder sb = new StringBuilder();
    while (n > 0) {
      sb.append(n % 2);
      n /= 2;
    }
    return sb.reverse().toString();
  }
}

8. 请实现一个字符串的反转。


public class ReverseString {
  public static String reverse(String s) {
    if (s == null) {
      return null;
    }
    return new StringBuilder(s).reverse().toString();
  }
}

9. 请实现一个快速排序算法。


public class QuickSort {
  public static void quickSort(int[] arr) {
    if (arr == null || arr.length <= 1) {
      return;
    }
    quickSort(arr, 0, arr.length - 1);
  }
  private static void quickSort(int[] arr, int left, int right) {
    if (left >= right) {
      return;
    }
    int pivot = partition(arr, left, right);
    quickSort(arr, left, pivot - 1);
    quickSort(arr, pivot + 1, right);
  }
  private static int partition(int[] arr, int left, int right) {
    int pivotValue = arr[right];
    int pivotIndex = left;
    for (int i = left; i < right; i++) {
      if (arr[i] < pivotValue) {
        swap(arr, i, pivotIndex);
        pivotIndex++;
      }
    }
    swap(arr, pivotIndex, right);
    return pivotIndex;
  }
  private static void swap(int[] arr, int i, int j) {
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
  }
}

10. 请实现一个图的深度优先遍历。


public class DFS {
  static class Graph {
    private int V;
    private LinkedList
  [] adj;
 
    public Graph(int V) {
      this.V = V;
      this.adj = new LinkedList[V];
      for (int i = 0; i < V; i++) {
        adj[i] = new LinkedList<>();
      }
    }
    public void addEdge(int v, int w) {
      adj[v].add(w);
    }
    public void DFS(int v) {
      boolean[] visited = new boolean[V];
      DFSUtil(v, visited);
    }
    private void DFSUtil(int v, boolean[] visited) {
      visited[v] = true;
      System.out.print(v + " ");
      Iterator
  it = adj[v].listIterator();
 
      while (it.hasNext()) {
        int n = it.next();
        if (!visited[n]) {
          DFSUtil(n, visited);
        }
      }
    }
  }
}

以上是本文分享的十份Java面试题及其代码案例，希望对广大面试党有所帮助。如果你想进一步了解Java知识或者其他编程语言知识，欢迎关注本公众号，获取更多优质文章和资源。

资源下载链接：百度网盘（XXXXXXX）

上一篇: idea打包java可执行jar包

下一篇: 如何使用Java实现图片的显示

评论区

()

相似文章