检查两个数组是否相等

给定两个给定的相等长度的数组，任务是查找给定的数组是否相等。如果两个数组都包含相同的元素集，则称两个数组相等，但是元素的排列(或排列)可能会不同。

注意：如果存在重复，则要使两个数组相等，重复元素的计数也必须相同。

例子 ：

Input  : arr1[] = {1, 2, 5, 4, 0};
         arr2[] = {2, 4, 5, 0, 1}; 
Output : Yes

Input  : arr1[] = {1, 2, 5, 4, 0, 2, 1};
         arr2[] = {2, 4, 5, 0, 1, 1, 2}; 
Output : Yes
 
Input : arr1[] = {1, 7, 1};
        arr2[] = {7, 7, 1};
Output : No

一个简单的解决方案是对两个数组进行排序，然后线性比较元素。

C++

// C++ program to find given two array
// are equal or not
#include 
using namespace std;
 
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
bool areEqual(int arr1[], int arr2[], int n, int m)
{
    // If lengths of array are not equal means
    // array are not equal
    if (n != m)
        return false;
 
    // Sort both arrays
    sort(arr1, arr1 + n);
    sort(arr2, arr2 + m);
 
    // Linearly compare elements
    for (int i = 0; i < n; i++)
        if (arr1[i] != arr2[i])
            return false;
 
    // If all elements were same.
    return true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 3, 5, 2, 5, 2 };
    int arr2[] = { 2, 3, 5, 5, 2 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
 
    if (areEqual(arr1, arr2, n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find given two array
// are equal or not
import java.io.*;
import java.util.*;
 
class GFG {
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    public static boolean areEqual(int arr1[], int arr2[])
    {
        int n = arr1.length;
        int m = arr2.length;
 
        // If lengths of array are not equal means
        // array are not equal
        if (n != m)
            return false;
 
        // Sort both arrays
        Arrays.sort(arr1);
        Arrays.sort(arr2);
 
        // Linearly compare elements
        for (int i = 0; i < n; i++)
            if (arr1[i] != arr2[i])
                return false;
 
        // If all elements were same.
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 3, 5, 2, 5, 2 };
        int arr2[] = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 program to find given
# two array are equal or not
 
# Returns true if arr1[0..n-1] and
# arr2[0..m-1] contain same elements.
 
 
def areEqual(arr1, arr2, n, m):
 
    # If lengths of array are not
    # equal means array are not equal
    if (n != m):
        return False
 
    # Sort both arrays
    arr1.sort()
    arr2.sort()
 
    # Linearly compare elements
    for i in range(0, n - 1):
        if (arr1[i] != arr2[i]):
            return False
 
    # If all elements were same.
    return True
 
 
# Driver Code
arr1 = [3, 5, 2, 5, 2]
arr2 = [2, 3, 5, 5, 2]
n = len(arr1)
m = len(arr2)
 
if (areEqual(arr1, arr2, n, m)):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Shivi_Aggarwal.

C#

// C# program to find given two array
// are equal or not
using System;
 
class GFG {
 
    // Returns true if arr1[0..n-1] and
    // arr2[0..m-1] contain same elements.
    public static bool areEqual(int[] arr1, int[] arr2)
    {
        int n = arr1.Length;
        int m = arr2.Length;
 
        // If lengths of array are not
        // equal means array are not equal
        if (n != m)
            return false;
 
        // Sort both arrays
        Array.Sort(arr1);
        Array.Sort(arr2);
 
        // Linearly compare elements
        for (int i = 0; i < n; i++)
            if (arr1[i] != arr2[i])
                return false;
 
        // If all elements were same.
        return true;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr1 = { 3, 5, 2, 5, 2 };
        int[] arr2 = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// JavaScript program for the above approach
 
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    function areEqual(arr1, arr2)
    {
        let n = arr1.length;
        let m = arr2.length;
 
        // If lengths of array are not equal means
        // array are not equal
        if (n != m)
            return false;
 
        // Sort both arrays
        arr1.sort();
        arr2.sort();
 
        // Linearly compare elements
        for (let i = 0; i < n; i++)
            if (arr1[i] != arr2[i])
                return false;
 
        // If all elements were same.
        return true;
    }
 
// Driver Code
    let arr1 = [ 3, 5, 2, 5, 2 ];
    let arr2 = [ 2, 3, 5, 5, 2 ];
 
    if (areEqual(arr1, arr2))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by chinmoy1997pal.

输出

Yes

时间复杂度： O(n log n)
辅助空间： O(1)

此方法的有效解决方案是使用哈希。我们将arr1 []的所有元素及其计数存储在哈希表中。然后我们遍历arr2 []并检查arr2 []中每个元素的计数是否与arr1 []中的计数匹配。

以下是上述想法的实现。我们使用unordered_map来存储计数。

C++

// C++ program to find given two array
// are equal or not using hashing technique
#include 
using namespace std;
 
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
bool areEqual(int arr1[], int arr2[], int n, int m)
{
    // If lengths of arrays are not equal
    if (n != m)
        return false;
 
    // Store arr1[] elements and their counts in
    // hash map
    unordered_map mp;
    for (int i = 0; i < n; i++)
        mp[arr1[i]]++;
 
    // Traverse arr2[] elements and check if all
    // elements of arr2[] are present same number
    // of times or not.
    for (int i = 0; i < n; i++) {
        // If there is an element in arr2[], but
        // not in arr1[]
        if (mp.find(arr2[i]) == mp.end())
            return false;
 
        // If an element of arr2[] appears more
        // times than it appears in arr1[]
        if (mp[arr2[i]] == 0)
            return false;
 
        mp[arr2[i]]--;
    }
 
    return true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 3, 5, 2, 5, 2 };
    int arr2[] = { 2, 3, 5, 5, 2 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
 
    if (areEqual(arr1, arr2, n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find given two array
// are equal or not using hashing technique
import java.util.*;
import java.io.*;
 
class GFG {
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    public static boolean areEqual(int arr1[], int arr2[])
    {
        int n = arr1.length;
        int m = arr2.length;
 
        // If lengths of arrays are not equal
        if (n != m)
            return false;
 
        // Store arr1[] elements and their counts in
        // hash map
        Map map
            = new HashMap();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (map.get(arr1[i]) == null)
                map.put(arr1[i], 1);
            else {
                count = map.get(arr1[i]);
                count++;
                map.put(arr1[i], count);
            }
        }
 
        // Traverse arr2[] elements and check if all
        // elements of arr2[] are present same number
        // of times or not.
        for (int i = 0; i < n; i++) {
            // If there is an element in arr2[], but
            // not in arr1[]
            if (!map.containsKey(arr2[i]))
                return false;
 
            // If an element of arr2[] appears more
            // times than it appears in arr1[]
            if (map.get(arr2[i]) == 0)
                return false;
 
            count = map.get(arr2[i]);
            --count;
            map.put(arr2[i], count);
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 3, 5, 2, 5, 2 };
        int arr2[] = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 program to find if given
# two arrays are equal or not
# using dictionary
from collections import defaultdict
 
# Returns true if arr1[0..n-1] and
# arr2[0..m-1] contain same elements.
 
 
def areEqual(arr1, arr2, n, m):
 
    # If lengths of array are not
    # equal means array are not equal
    if (n != m):
        return False
 
    # Create a defaultdict count to
    # store counts
    count = defaultdict(int)
 
    # Store the elements of arr1
    # and their counts in the dictionary
    for i in arr1:
        count[i] += 1
 
    # Traverse through arr2 and compare
    # the elements and its count with
    # the elements of arr1
    for i in arr2:
 
        # Return false if the elemnent
        # is not in arr2 or if any element
        # appears more no. of times than in arr1
        if (count[i] == 0):
            return False
 
        # If element is found, decrement
        # its value in the dictionary
        else:
            count[i] -= 1
 
    # Return true if both arr1 and
    # arr2 are equal
    return True
 
 
# Driver Code
arr1 = [3, 5, 2, 5, 2]
arr2 = [2, 3, 5, 5, 2]
 
n = len(arr1)
m = len(arr2)
 
if areEqual(arr1, arr2, n, m):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Karthik_Aravind

C＃

// C# program to find given two array
// are equal or not using hashing technique
using System;
using System.Collections.Generic;
 
class GFG {
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    public static bool areEqual(int[] arr1, int[] arr2)
    {
        int n = arr1.Length;
        int m = arr2.Length;
 
        // If lengths of arrays are not equal
        if (n != m)
            return false;
 
        // Store arr1[] elements and their counts in
        // hash map
        Dictionary map
            = new Dictionary();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (!map.ContainsKey(arr1[i]))
                map.Add(arr1[i], 1);
            else {
                count = map[arr1[i]];
                count++;
                map.Remove(arr1[i]);
                map.Add(arr1[i], count);
            }
        }
 
        // Traverse arr2[] elements and check if all
        // elements of arr2[] are present same number
        // of times or not.
        for (int i = 0; i < n; i++) {
            // If there is an element in arr2[], but
            // not in arr1[]
            if (!map.ContainsKey(arr2[i]))
                return false;
 
            // If an element of arr2[] appears more
            // times than it appears in arr1[]
            if (map[arr2[i]] == 0)
                return false;
 
            count = map[arr2[i]];
            --count;
 
            if (!map.ContainsKey(arr2[i]))
                map.Add(arr2[i], count);
        }
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr1 = { 3, 5, 2, 5, 2 };
        int[] arr2 = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
/* This code contributed by PrinciRaj1992 */

输出

Yes

时间复杂度： O(n)
辅助空间： O(n)

一种不对数组的每个元素进行比较且不使用unordered_map(通过使用XOR)的替代解决方案。仅当每个元素在数组中仅存在一次时，此方法才有效。例如：数组a：{3，3}和数组b：{5，5}，xor_of_array_a(例如b1)= 0和xor_of_array_b = 0(例如b2)和b1 ^ b2 = 0，但是数组a和数组b是不相等。

C++ 14

// C++ program to find given two array
// are equal or not
#include 
using namespace std;
 
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
bool areEqual(int arr1[], int arr2[], int n, int m)
{
    // If lengths of array are not equal means
    // array are not equal
    if (n != m)
        return false;
     
    // to store xor of both arrays
    int b1 = arr1[0];
    int b2 = arr2[0];
     
    // find xor of each elements in array
    for (int i = 1; i < n; i++) {
        b1 ^= arr1[i];
    }
 
    for (int i = 1; i < m; i++) {
        b2 ^= arr2[i];
    }
    int all_xor = b1 ^ b2;
     
    // if xor is zero means they are equal (5^5=0)
    if (all_xor == 0)
        return true;
     
    // If all elements were not same, then xor will not be
    // zero
    return false;
}
 
// Driver Code
int main()
{
    int arr1[] = { 3, 6, 7, 5, 2 };
    int arr2[] = { 2, 3, 5, 6, 7 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
 
    // Function call
    if (areEqual(arr1, arr2, n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find given two array
// are equal or not
import java.io.*;
import java.util.*;
 
class GFG{
     
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
public static boolean areEqual(int arr1[],
                               int arr2[])
{
     
    // Length of the two array
    int n = arr1.length;
    int m = arr2.length;
     
    // If lengths of arrays are not equal
    if (n != m)
        return false;
         
    // To store xor of both arrays
    int b1 = arr1[0];
    int b2 = arr2[0];
     
    // Find xor of each elements in array
    for(int i = 1; i < n; i++)
    {
        b1 ^= arr1[i];
    }
    for(int i = 1; i < m; i++)
    {
        b2 ^= arr2[i];
    }
    int all_xor = b1 ^ b2;
     
    // If xor is zero means they are
    // equal (5^5=0)
    if (all_xor == 0)
        return true;
         
    // If all elements were not same,
    // then xor will not be zero
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    int arr1[] = { 3, 5, 2, 5, 2 };
    int arr2[] = { 2, 3, 5, 5, 2 };
     
    // Function call
    if (areEqual(arr1, arr2))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by sayantanbose2001

Python3

# Python3 program to find given
# two array are equal or not
 
# Returns true if arr1[0..n-1] and
# arr2[0..m-1] contain same elements.
 
def areEqual(arr1, arr2, n, m):
 
    # If lengths of array are not
    # equal means array are not equal
    if (n != m):
        return False
    b1 = arr1[0]
    b2 = arr2[0]
     
    # find xor of all elements
    for i in range(1, n - 1):
        b1 ^= arr1[i]
 
    for i in range(1, m - 1):
        b2 ^= arr2[i]
 
    all_xor = b1 ^ b2
     
    # If all elements were same then xor will be zero
    if(all_xor == 0):
        return True
 
    return False
 
 
# Driver Code
arr1 = [3, 5, 2, 5, 2]
arr2 = [2, 3, 5, 5, 2]
n = len(arr1)
m = len(arr2)
 
# Function call
if (areEqual(arr1, arr2, n, m)):
    print("Yes")
else:
    print("No")

输出

Yes

时间复杂度： O(n)  
辅助空间： O(1)

改进者： vt_m，Shivi_Aggarwal，imrohan，princiraj1992，karthikaravindt88， anupriyanishad