如何检查两个给定的集合是否不相交？

给定两个数组表示的两个集合，如何检查给定的两个集合是否不相交？可以假设给定的数组没有重复项。

Input: set1[] = {12, 34, 11, 9, 3}
       set2[] = {2, 1, 3, 5}
Output: Not Disjoint
3 is common in two sets.

Input: set1[] = {12, 34, 11, 9, 3}
       set2[] = {7, 2, 1, 5}
Output: Yes, Disjoint
There is no common element in two sets.

有很多方法可以解决这个问题，这是一个很好的测试，看看你能猜到多少个解决方案。

方法一（简单）
遍历第一个集合的每个元素并在另一个集合中搜索它，如果找到任何元素，则返回false。如果没有找到元素，则返回树。该方法的时间复杂度为 O(mn)。

下面是上述想法的实现。

C++

// A Simple C++ program to check if two sets are disjoint
#include
using namespace std;
 
// Returns true if set1[] and set2[] are disjoint, else false
bool areDisjoint(int set1[], int set2[], int m, int n)
{
    // Take every element of set1[] and search it in set2
    for (int i=0; i

---

Java

// Java program to check if two sets are disjoint
 
public class disjoint1
{
    // Returns true if set1[] and set2[] are
    // disjoint, else false
    boolean aredisjoint(int set1[], int set2[])
    {
         // Take every element of set1[] and
         // search it in set2
        for (int i = 0; i < set1.length; i++)
        {
            for (int j = 0; j < set2.length; j++)
            {
                if (set1[i] == set2[j])
                    return false;
            }
        }
        // If no element of set1 is present in set2
        return true;
    }
     
    // Driver program to test above function
    public static void main(String[] args)
    {
        disjoint1 dis = new disjoint1();
        int set1[] = { 12, 34, 11, 9, 3 };
        int set2[] = { 7, 2, 1, 5 };
 
        boolean result = dis.aredisjoint(set1, set2);
        if (result)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Rishabh Mahrsee

Python

# A Simple python 3 program to check
# if two sets are disjoint
 
# Returns true if set1[] and set2[] are disjoint, else false
def areDisjoint(set1, set2, m, n):
    # Take every element of set1[] and search it in set2
    for i in range(0, m):
        for j in range(0, n):
            if (set1[i] == set2[j]):
                return False
 
    # If no element of set1 is present in set2
    return True
 
 
# Driver program
set1 = [12, 34, 11, 9, 3]
set2 = [7, 2, 1, 5]
m = len(set1)
n = len(set2)
print("yes") if areDisjoint(set1, set2, m, n) else(" No")
 
# This code ia contributed by Smitha Dinesh Semwal

C#

// C# program to check if two
// sets are disjoint
using System;
 
class GFG
{
// Returns true if set1[] and set2[]
// are disjoint, else false
public virtual bool aredisjoint(int[] set1,
                                int[] set2)
{
    // Take every element of set1[]
    // and search it in set2
    for (int i = 0; i < set1.Length; i++)
    {
        for (int j = 0;
                 j < set2.Length; j++)
        {
            if (set1[i] == set2[j])
            {
                return false;
            }
        }
    }
     
    // If no element of set1 is
    // present in set2
    return true;
}
 
// Driver Code
public static void Main(string[] args)
{
    GFG dis = new GFG();
    int[] set1 = new int[] {12, 34, 11, 9, 3};
    int[] set2 = new int[] {7, 2, 1, 5};
 
    bool result = dis.aredisjoint(set1, set2);
    if (result)
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by Shrikant13

Javascript

 
// Javascript program to check if two sets are disjoint   
     
    // Returns true if set1[] and set2[] are
    // disjoint, else false
    function aredisjoint(set1,set2)
    {
        // Take every element of set1[] and
         // search it in set2
        for (let i = 0; i < set1.length; i++)
        {
            for (let j = 0; j < set2.length; j++)
            {
                if (set1[i] == set2[j])
                    return false;
            }
        }
        // If no element of set1 is present in set2
        return true;
    }
     
    // Driver program to test above function
    let set1=[12, 34, 11, 9, 3];
    let set2=[7, 2, 1, 5];
    result = aredisjoint(set1, set2);
    if (result)
        document.write("Yes");
    else
        document.write("No");
     
    // This code is contributed by avanitrachhadiya2155
     

---

输出 ：

Yes

方法二（使用排序合并）
1) 对第一组和第二组进行排序。
2）使用像过程一样的合并来比较元素。

下面是上述想法的实现。

C++

// A Simple C++ program to check if two sets are disjoint
#include
using namespace std;
 
// Returns true if set1[] and set2[] are disjoint, else false
bool areDisjoint(int set1[], int set2[], int m, int n)
{
    // Sort the given two sets
    sort(set1, set1+m);
    sort(set2, set2+n);
 
    // Check for same elements using merge like process
    int i = 0, j = 0;
    while (i < m && j < n)
    {
        if (set1[i] < set2[j])
            i++;
        else if (set2[j] < set1[i])
            j++;
        else /* if set1[i] == set2[j] */
            return false;
    }
 
    return true;
}
 
// Driver program to test above function
int main()
{
    int set1[] = {12, 34, 11, 9, 3};
    int set2[] = {7, 2, 1, 5};
    int m = sizeof(set1)/sizeof(set1[0]);
    int n = sizeof(set2)/sizeof(set2[0]);
    areDisjoint(set1, set2, m, n)? cout << "Yes" : cout << " No";
    return 0;
}

Java

// Java program to check if two sets are disjoint
 
import java.util.Arrays;
 
public class disjoint1
{
    // Returns true if set1[] and set2[] are
    // disjoint, else false
    boolean aredisjoint(int set1[], int set2[])
    {
        int i=0,j=0;
         
        // Sort the given two sets
        Arrays.sort(set1);
        Arrays.sort(set2);
         
        // Check for same elements using
        // merge like process
        while(iset2[j])
                j++;
            else
                return false;
             
        }
        return true;
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        disjoint1 dis = new disjoint1();
        int set1[] = { 12, 34, 11, 9, 3 };
        int set2[] = { 7, 2, 1, 5 };
 
        boolean result = dis.aredisjoint(set1, set2);
        if (result)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
 
// This code is contributed by Rishabh Mahrsee

蟒蛇3

# A Simple Python 3 program to check
# if two sets are disjoint
 
# Returns true if set1[] and set2[]
# are disjoint, else false
def areDisjoint(set1, set2, m, n):
    # Sort the given two sets
    set1.sort()
    set2.sort()
 
    # Check for same elements 
    # using merge like process
    i = 0; j = 0
    while (i < m and j < n):
         
        if (set1[i] < set2[j]):
            i += 1
        elif (set2[j] < set1[i]):
            j += 1
        else: # if set1[i] == set2[j]
            return False
    return True
 
 
# Driver Code
set1 = [12, 34, 11, 9, 3]
set2 = [7, 2, 1, 5]
m = len(set1)
n = len(set2)
 
print("Yes") if areDisjoint(set1, set2, m, n) else print("No")
 
# This code is contributed by Smitha Dinesh Semwal

C＃

// C# program to check if two sets are disjoint
using System;
 
public class disjoint1
{
    // Returns true if set1[] and set2[] are
    // disjoint, else false
    Boolean aredisjoint(int []set1, int []set2)
    {
        int i = 0, j = 0;
         
        // Sort the given two sets
        Array.Sort(set1);
        Array.Sort(set2);
         
        // Check for same elements using
        // merge like process
        while(i < set1.Length && j < set2.Length)
        {
            if(set1[i] < set2[j])
                i++;
            else if(set1[i] > set2[j])
                j++;
            else
                return false;
             
        }
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        disjoint1 dis = new disjoint1();
        int []set1 = { 12, 34, 11, 9, 3 };
        int []set2 = { 7, 2, 1, 5 };
 
        Boolean result = dis.aredisjoint(set1, set2);
        if (result)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript program to check if two
// sets are disjoint
 
// Returns true if set1[] and set2[] are
// disjoint, else false
function aredisjoint(set1, set2)
{
    let i = 0, j = 0;
     
    // Sort the given two sets
    set1.sort(function(a, b){return a - b});
    set2.sort(function(a, b){return a - b});
     
    // Check for same elements using
    // merge like process
    while (i < set1.length && j < set2.length)
    {
        if (set1[i] < set2[j])
            i++;
        else if (set1[i] > set2[j])
            j++;
        else
            return false;
    }
    return true;
}
 
// Driver code
let set1 = [ 12, 34, 11, 9, 3 ];
let set2 = [ 7, 2, 1, 5 ];
result = aredisjoint(set1, set2);
 
if (result)
    document.write("YES");
else
    document.write("NO");
 
// This code is contributed by rag2127
 

输出 ：

Yes

上述解决方案的时间复杂度为 O(mLogm + nLogn)。
上述解决方案首先对两个集合进行排序，然后花费 O(m+n) 时间来找到交集。如果给定输入集已排序，那么这种方法是最好的。

方法三（使用排序和二分查找）
这类似于方法 1。我们使用二分搜索代替线性搜索。
1) 排序第一组。
2) 遍历第二个集合中的每一个元素，并使用二分查找搜索第一个集合中的每个元素。如果找到一个元素，则返回它。
该方法的时间复杂度为 O(mLogm + nLogm)

方法四（使用二叉搜索树）
1）创建第一组所有元素的自平衡二叉搜索树（Red Black、AVL、Splay等）。
2) 遍历第二个集合的所有元素，并在上面构造的二叉搜索树中搜索每个元素。如果找到该元素，则返回 false。
3) 如果所有元素都不存在，则返回真。
该方法的时间复杂度为 O(mLogm + nLogm)。

方法 5（使用哈希）
1) 创建一个空的哈希表。
2) 遍历第一个集合并将每个元素存储在哈希表中。
3) 遍历第二组并检查哈希表中是否存在任何元素。如果存在，则返回 false，否则忽略该元素。
4) 如果哈希表中不存在第二个集合的所有元素，则返回真。

下面是这个方法的实现。

C++

/* C++ program to check if two sets are distinct or not */
#include
using namespace std;
 
// This function prints all distinct elements
bool areDisjoint(int set1[], int set2[], int n1, int n2)
{
    // Creates an empty hashset
    set myset;
 
    // Traverse the first set and store its elements in hash
    for (int i = 0; i < n1; i++)
        myset.insert(set1[i]);
 
    // Traverse the second set and check if any element of it
    // is already in hash or not.
    for (int i = 0; i < n2; i++)
        if (myset.find(set2[i]) != myset.end())
            return false;
 
    return true;
}
 
// Driver method to test above method
int main()
{
    int set1[] = {10, 5, 3, 4, 6};
    int set2[] = {8, 7, 9, 3};
 
    int n1 = sizeof(set1) / sizeof(set1[0]);
    int n2 = sizeof(set2) / sizeof(set2[0]);
    if (areDisjoint(set1, set2, n1, n2))
        cout << "Yes";
    else
        cout << "No";
}
//This article is contributed by Chhavi

Java

/* Java program to check if two sets are distinct or not */
import java.util.*;
 
class Main
{
    // This function prints all distinct elements
    static boolean areDisjoint(int set1[], int set2[])
    {
        // Creates an empty hashset
        HashSet set = new HashSet<>();
 
        // Traverse the first set and store its elements in hash
        for (int i=0; i

蟒蛇3

# Python3 program to
# check if two sets are
# distinct or not
# This function prints
# all distinct elements
def areDisjoint(set1, set2,
                n1, n2):
   
  # Creates an empty hashset
  myset = set([])
   
  # Traverse the first set
  # and store its elements in hash
  for i in range (n1):
    myset.add(set1[i])
     
  # Traverse the second set
  # and check if any element of it
  # is already in hash or not.
  for i in range (n2):
    if (set2[i] in myset):
      return False
  return True
 
# Driver method to test above method
if __name__ == "__main__":
   
  set1 = [10, 5, 3, 4, 6]
  set2 = [8, 7, 9, 3]
 
  n1 = len(set1)
  n2 = len(set2)
   
  if (areDisjoint(set1, set2,
                  n1, n2)):
    print ("Yes")
  else:
    print("No")
 
# This code is contributed by Chitranayal

C＃

using System;
using System.Collections.Generic;
 
/* C# program to check if two sets are distinct or not */
 
public class GFG
{
    // This function prints all distinct elements
    public static bool areDisjoint(int[] set1, int[] set2)
    {
        // Creates an empty hashset
        HashSet set = new HashSet();
 
        // Traverse the first set and store its elements in hash
        for (int i = 0; i < set1.Length; i++)
        {
            set.Add(set1[i]);
        }
 
        // Traverse the second set and check if any element of it
        // is already in hash or not.
        for (int i = 0; i < set2.Length; i++)
        {
            if (set.Contains(set2[i]))
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver method to test above method
    public static void Main(string[] args)
    {
        int[] set1 = new int[] {10, 5, 3, 4, 6};
        int[] set2 = new int[] {8, 7, 9, 3};
        if (areDisjoint(set1, set2))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
//This code is contributed by Shrikant13

Javascript

/* Javascript program to check if two sets are distinct or not */
 
 // This function prints all distinct elements
function areDisjoint(set1,set2)
{
    // Creates an empty hashset
        let set = new Set();
  
        // Traverse the first set and store its elements in hash
        for (let i = 0; i < set1.length; i++)
            set.add(set1[i]);
  
        // Traverse the second set and check if any element of it
        // is already in hash or not.
        for (let i = 0; i < set2.length; i++)
            if (set.has(set2[i]))
                return false;
  
        return true;
}
 
// Driver method to test above method
let set1 = [10, 5, 3, 4, 6];
let set2 = [8, 7, 9, 3];
if (areDisjoint(set1, set2))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by ab2127

输出：

No

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