在两个数组之间查找公共元素的Java程序

给定两个数组，我们的任务是找到它们的公共元素。
例子：

Input:  Array1 = ["Article", "for", "Geeks", "for", "Geeks"], 
        Array2 = ["Article", "Geeks", "Geeks"]
Output: [Article,Geeks]

Input:  Array1 = ["a", "b", "c", "d", "e", "f"], 
        Array2 = ["b", "d", "e", "h", "g", "c"]
Output: [b, c, d, e]

使用迭代方法

方法：

1. 获取两个Java数组。
2. 逐个遍历数组的每个元素并检查它们是否在两者中通用。
3. 为唯一条目添加集合中的每个公共元素。

Java

// Java Program to find common elements
// in two Arrays
// Using iterative method
 
import java.io.*;
import java.util.*;
 
class GFG {
    private static void FindCommonElemet(String[] arr1,
                                         String[] arr2)
    {
        Set set = new HashSet<>();
        for (int i = 0; i < arr1.length; i++) {
            for (int j = 0; j < arr2.length; j++) {
                if (arr1[i] == arr2[j]) {
 
                    // add common elements
                    set.add(arr1[i]);
                    break;
                }
            }
        }
        for (String i : set) {
            System.out.print(i + " ");
        }
    }
 
    // main method
    public static void main(String[] args)
    {
 
        // create Array 1
        String[] arr1
            = { "Article", "in", "Geeks", "for", "Geeks" };
 
        // create Array 2
        String[] arr2 = { "Geeks", "for", "Geeks" };
 
        // print Array 1
        System.out.println("Array 1: "
                           + Arrays.toString(arr1));
 
        // print Array 2
        System.out.println("Array 2: "
                           + Arrays.toString(arr2));
 
        System.out.print("Common Elements: ");
 
        // Find the common elements
        FindCommonElemet(arr1, arr2);
    }
}

输出

Array 1: [Article, in, Geeks, for, Geeks]
Array 2: [Geeks, for, Geeks]
Common Elements: Geeks for 

时间复杂度： O(n^2)

使用哈希集：

通过使用HashSet的retainAll()方法，我们可以找到两个数组之间的公共元素。

句法：

// This method keeps only the common elements
// of both Collection in Collection1.

Collections1.retainAll(Collections2)

方法 ：

1. 获取两个数组。
2. 创建两个散列集并从这些集的数组中添加元素。
3. 使用Collection.retainAll() 方法查找两个集合中的公共元素。此方法仅保留 Collection1 中两个 Collection 的公共元素。
4. 集合 1 现在仅包含公共元素。

下面是上述方法的实现：

Java

// Java Program to find common elements
// in two Arrays using hashsets
// and retainAll() method
import java.io.*;
import java.util.*;
 
class GFG {
 
    // function to create hashsets
    // from arrays and find
    // their common element
    public static void FindCommonElements(int[] arr1,
                                          int[] arr2)
    {
        // create hashsets
        Set set1 = new HashSet<>();
        Set set2 = new HashSet<>();
 
        // Adding elements from array1
        for (int i : arr1) {
            set1.add(i);
        }
 
        // Adding elements from array2
        for (int i : arr2) {
            set2.add(i);
        }
 
        // use retainAll() method to
        // find common elements
        set1.retainAll(set2);
        System.out.println("Common elements- " + set1);
    }
 
    // main method
    public static void main(String[] args)
    {
        // create Array 1
        int[] arr1
            = { 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 };
 
        // create Array 2
        int[] arr2 = { 100, 9, 64, 7, 36, 5, 16, 3, 4, 1 };
 
        // print Array 1
        System.out.println("Array 1: "
                           + Arrays.toString(arr1));
        // print Array 2
        System.out.println("Array 2: "
                           + Arrays.toString(arr2));
        FindCommonElements(arr1, arr2);
    }
}

输出

Array 1: [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
Array 2: [100, 9, 64, 7, 36, 5, 16, 3, 4, 1]
Common elements- [16, 64, 1, 4, 36, 100, 9]

时间复杂度： O(n)                                                                                                                                   
使用哈希集：

方法：

1. 将第一个数组的所有元素添加到哈希集中。

2. 迭代第二个数组并使用 contains 方法检查 hashset 中是否存在元素。如果 contains == true，则添加元素以生成数组。

下面是上述方法的实现：

Java

// Java program for the above approach
import java.io.*;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
 
class Test {
 
    private static void findCommonElements(int[] arr1,
                                           int[] arr2)
    {
         
        // Check if length of arr1 is greater than 0
        // and length of arr2 is greater than 0
        if (arr1.length > 0 && arr2.length > 0) {
            Set firstSet = new HashSet();
            for (int i = 0; i < arr1.length; i++) {
                firstSet.add(arr1[i]);
            }
           
            // Iterate the elements of the arr2
            for (int j = 0; j < arr2.length; j++) {
                if (firstSet.contains(arr2[j])) {
                    System.out.println(arr2[j]);
                }
            }
        }
    }
   
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6, 7 };
        int[] arr2 = new int[] { 1, 3, 4, 5, 6, 9, 8 };
       
        // Function Call
        findCommonElements(arr1, arr2);
    }
}

输出

1
3
4
5
6

时间复杂度：O(n)