给定两个长度相等的给定数组，任务是找出给定数组是否相等。如果两个数组都包含相同的元素集，则称其为相等，但元素的排列（或排列）可能不同。
null
注： 如果存在重复，则重复元素的计数也必须相同，才能使两个数组相等。
例如：
Input : arr1[] = {1, 2, 5, 4, 0}; arr2[] = {2, 4, 5, 0, 1}; Output : YesInput : 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
A. 简单解决方案 就是对两个数组进行排序，然后对元素进行线性比较。
C++
// C++ program to find given two array // are equal or not #include <bits/stdc++.h> 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. |
PHP
<?php // PHP 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. function 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 sort( $arr1 ); sort( $arr2 ); // Linearly compare elements for ( $i = 0; $i < $n ; $i ++) if ( $arr1 [ $i ] != $arr2 [ $i ]) return false; // If all elements were same. return true; } // Driver Code $arr1 = array ( 3, 5, 2, 5, 2); $arr2 = array ( 2, 3, 5, 5, 2); $n = count ( $arr1 ); $m = count ( $arr2 ); if (areEqual( $arr1 , $arr2 , $n , $m )) echo "Yes" ; else echo "No" ; // This code is contributed by anuj_67. ?> |
Javascript
<script> // 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. </script> |
输出
Yes
时间复杂性： O（n日志n） 辅助空间： O（1）
一 有效解决方案 这种方法的关键是使用哈希。我们将arr1[]的所有元素及其计数存储在一个哈希表中。然后遍历arr2[]并检查arr2[]中每个元素的计数是否与arr1[]中的计数匹配。
下面是上述想法的实施。我们使用 无序地图 存储计数。
C++
// C++ program to find given two array // are equal or not using hashing technique #include <bits/stdc++.h> 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< int , int > 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<Integer, Integer> map = new HashMap<Integer, Integer>(); 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 element # 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< int , int > map = new Dictionary< int , int >(); 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 */ |
Javascript
<script> // Javascript program to find given two array // are equal or not using hashing technique // 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 arrays are not equal if (n != m) return false ; // Store arr1[] elements and their counts in // hash map let map = new Map(); let count = 0; for (let i = 0; i < n; i++) { if (map.get(arr1[i]) == null ) map.set(arr1[i], 1); else { count = map.get(arr1[i]); count++; map.set(arr1[i], count); } } // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (let i = 0; i < n; i++) { // If there is an element in arr2[], but // not in arr1[] if (!map.has(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.set(arr2[i], count); } 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" ); </script> |
输出
Yes
时间复杂性： O（n） 辅助空间： O（n）
一 替代方案 不比较数组的每个元素，也不使用无序的_映射（通过使用异或）。只有当每个元素在一个数组中只存在一次时，这种方法才会起作用。例如：数组a:{3,3}和数组b:{5,5}，数组a的xor_（假设b1）=0和数组b的xor_=0（假设b2）和b1^b2=0，但数组a和数组b不相等。
C++14
// C++ program to find given two array // are equal or not #include <bits/stdc++.h> 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" ) |
Javascript
<script> // Java Script 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. function areEqual(arr1,arr2) { // Length of the two array let n = arr1.length; let m = arr2.length; // If lengths of arrays are not equal if (n != m) return false ; // To store xor of both arrays let b1 = arr1[0]; let b2 = arr2[0]; // Find xor of each elements in array for (let i = 1; i < n; i++) { b1 ^= arr1[i]; } for (let i = 1; i < m; i++) { b2 ^= arr2[i]; } let 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 let arr1 = [ 3, 5, 2, 5, 2 ]; let arr2 = [ 2, 3, 5, 5, 2 ]; // Function call if (areEqual(arr1, arr2)) document.write( "Yes" ); else document.write( "No" ); //contributed by sravan kumar Gottumukkala </script> |
输出
Yes
时间复杂性： O（n） 辅助空间： O（1）
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