给定两个不同元素的未排序数组，任务是从两个数组中查找其和等于的所有对 十、 . 例如：
null
Input : arr1[] = {-1, -2, 4, -6, 5, 7} arr2[] = {6, 3, 4, 0} x = 8Output : 4 4 5 3Input : arr1[] = {1, 2, 4, 5, 7} arr2[] = {5, 6, 3, 4, 8} x = 9Output : 1 8 4 5 5 4
请进来 : 亚马逊
A. 天真的方法 就是简单地运行两个循环并从两个数组中选取元素。逐个检查两个元素之和是否等于给定值x。
C++
// C++ program to find all pairs in both arrays // whose sum is equal to given value x #include <bits/stdc++.h> using namespace std; // Function to print all pairs in both arrays // whose sum is equal to given value x void findPairs( int arr1[], int arr2[], int n, int m, int x) { for ( int i = 0; i < n; i++) for ( int j = 0; j < m; j++) if (arr1[i] + arr2[j] == x) cout << arr1[i] << " " << arr2[j] << endl; } // Driver code int main() { int arr1[] = { 1, 2, 3, 7, 5, 4 }; int arr2[] = { 0, 7, 4, 3, 2, 1 }; int n = sizeof (arr1) / sizeof ( int ); int m = sizeof (arr2) / sizeof ( int ); int x = 8; findPairs(arr1, arr2, n, m, x); return 0; } |
JAVA
// Java program to find all pairs in both arrays // whose sum is equal to given value x import java.io.*; class GFG { // Function to print all pairs in both arrays // whose sum is equal to given value x static void findPairs( int arr1[], int arr2[], int n, int m, int x) { for ( int i = 0 ; i < n; i++) for ( int j = 0 ; j < m; j++) if (arr1[i] + arr2[j] == x) System.out.println(arr1[i] + " " + arr2[j]); } // Driver code public static void main(String[] args) { int arr1[] = { 1 , 2 , 3 , 7 , 5 , 4 }; int arr2[] = { 0 , 7 , 4 , 3 , 2 , 1 }; int x = 8 ; findPairs(arr1, arr2, arr1.length, arr2.length, x); } } // This code is contributed // by sunnysingh |
Python3
# Python 3 program to find all # pairs in both arrays whose # sum is equal to given value x # Function to print all pairs # in both arrays whose sum is # equal to given value x def findPairs(arr1, arr2, n, m, x): for i in range ( 0 , n): for j in range ( 0 , m): if (arr1[i] + arr2[j] = = x): print (arr1[i], arr2[j]) # Driver code arr1 = [ 1 , 2 , 3 , 7 , 5 , 4 ] arr2 = [ 0 , 7 , 4 , 3 , 2 , 1 ] n = len (arr1) m = len (arr2) x = 8 findPairs(arr1, arr2, n, m, x) # This code is contributed by Smitha Dinesh Semwal |
C#
// C# program to find all // pairs in both arrays // whose sum is equal to // given value x using System; class GFG { // Function to print all // pairs in both arrays // whose sum is equal to // given value x static void findPairs( int [] arr1, int [] arr2, int n, int m, int x) { for ( int i = 0; i < n; i++) for ( int j = 0; j < m; j++) if (arr1[i] + arr2[j] == x) Console.WriteLine(arr1[i] + " " + arr2[j]); } // Driver code static void Main() { int [] arr1 = { 1, 2, 3, 7, 5, 4 }; int [] arr2 = { 0, 7, 4, 3, 2, 1 }; int x = 8; findPairs(arr1, arr2, arr1.Length, arr2.Length, x); } } // This code is contributed // by Sam007 |
PHP
<?php // PHP program to find all pairs // in both arrays whose sum is // equal to given value x // Function to print all pairs // in both arrays whose sum is // equal to given value x function findPairs( $arr1 , $arr2 , $n , $m , $x ) { for ( $i = 0; $i < $n ; $i ++) for ( $j = 0; $j < $m ; $j ++) if ( $arr1 [ $i ] + $arr2 [ $j ] == $x ) echo $arr1 [ $i ] . " " . $arr2 [ $j ] . "" ; } // Driver code $arr1 = array (1, 2, 3, 7, 5, 4); $arr2 = array (0, 7, 4, 3, 2, 1); $n = count ( $arr1 ); $m = count ( $arr2 ); $x = 8; findPairs( $arr1 , $arr2 , $n , $m , $x ); // This code is contributed // by Sam007 ?> |
Javascript
<script> // Javascript program to find all pairs in both arrays // whose sum is equal to given value x // Function to print all pairs in both arrays // whose sum is equal to given value x function findPairs(arr1, arr2, n, m, x) { for (let i = 0; i < n; i++) for (let j = 0; j < m; j++) if (arr1[i] + arr2[j] == x) document.write(arr1[i] + " " + arr2[j] + "<br>" ); } // Driver code let arr1 = [ 1, 2, 3, 7, 5, 4 ]; let arr2 = [ 0, 7, 4, 3, 2, 1 ]; let n = arr1.length; let m = arr2.length; let x = 8; findPairs(arr1, arr2, n, m, x); // This code is contributed by Surbhi Tyagi. </script> |
输出：
1 77 15 34 4
时间复杂性： O（n^2） 辅助空间： O（1）
搜索方法 ：正如我们所知，排序算法可以在O（n logn）时间内对数据进行排序。因此，我们将选择O（n logn）时间算法，如：快速排序或堆排序。对于第二个数组的每个元素，我们将从K中减去它，并在第一个数组中搜索它。
步骤：
- 首先使用O（n logn）算法对给定数组进行排序，比如堆排序或快速排序。
- 为array-B（0到n）的每个元素运行一个循环。
- 在循环内部，使用临时变量temp，temp=K–B[i]。
- 使用二进制搜索（log n）在第一个数组即A中搜索temp变量。
如果元素是在A中找到的，则存在A∈ A和b∈ 使得a+B=K。
代码：
C++
#include<bits/stdc++.h> using namespace std; void heapify( int a[] , int n , int i) { int rootLargest = i; int lchild = 2 * i; int rchild = (2 * i) + 1; if (lchild < n && a[lchild] > a[rootLargest]) rootLargest = lchild; if (rchild < n && a[rchild] > a[rootLargest]) rootLargest = rchild; if (rootLargest != i) { swap(a[i] , a[rootLargest]); //Recursion heapify(a , n , rootLargest); } } int binarySearch( int a[] , int l , int r , int x) { while (l <= r) { int m = l + (r - l) / 2; if (a[m] == x) return m; if (a[m] < x) l = m + 1; else r = m - 1; } return -1; } int main() { int A[] = {1,2,1,3,4}; int B[] = {3,1,5,1,2}; int K = 8; int n = sizeof (A) / sizeof (A[0]); // Building the heap for ( int i = n / 2 - 1 ; i >= 1; i--) heapify(A , n , i); for ( int i=0 ; i<n ; i++) //O(n) { int temp = K - B[i]; //O(1) if (binarySearch(A , 0 , n-1 , temp)) //O(logn) { cout<< "Found the elements." ; break ; } } return 0; } |
JAVA
import java.util.*; class GFG{ static int A[] = { 1 , 2 , 1 , 3 , 4 }; static void heapify( int n , int i) { int rootLargest = i; int lchild = 2 * i; int rchild = ( 2 * i) + 1 ; if (lchild < n && A[lchild] > A[rootLargest]) rootLargest = lchild; if (rchild < n && A[rchild] > A[rootLargest]) rootLargest = rchild; if (rootLargest != i) { int t = A[i]; A[i] = A[rootLargest]; A[rootLargest] = t; //Recursion heapify( n , rootLargest); } } static int binarySearch( int l , int r , int x) { while (l <= r) { int m = l + (r - l) / 2 ; if (A[m] == x) return m; if (A[m] < x) l = m + 1 ; else r = m - 1 ; } return - 1 ; } public static void main(String[] args) { int B[] = { 3 , 1 , 5 , 1 , 2 }; int K = 8 ; int n = A.length; // Building the heap for ( int i = n / 2 - 1 ; i >= 1 ; i--) heapify( n , i); for ( int i = 0 ; i < n; i++) //O(n) { int temp = K - B[i]; //O(1) if (binarySearch( 0 , n - 1 , temp - 1 ) != - 1 ) //O(logn) { System.out.print( "Found the elements." ); break ; } } } } // This code is contributed by Rajput-Ji |
Python3
A = [ 1 , 2 , 1 , 3 , 4 ]; def heapify(n, i): rootLargest = i; lchild = 2 * i; rchild = ( 2 * i) + 1 ; if (lchild < n and A[lchild] > A[rootLargest]): rootLargest = lchild; if (rchild < n and A[rchild] > A[rootLargest]): rootLargest = rchild; if (rootLargest ! = i): t = A[i]; A[i] = A[rootLargest]; A[rootLargest] = t; # Recursion heapify(n, rootLargest); def binarySearch(l, r, x): while (l < = r): m = l + (r - l) / / 2 ; if (A[m] = = x): return m; if (A[m] < x): l = m + 1 ; else : r = m - 1 ; return - 1 ; if __name__ = = '__main__' : B = [ 3 , 1 , 5 , 1 , 2 ]; K = 8 ; n = len (A); # Building the heap for i in range (n / / 2 - 1 , 0 , - 1 ): heapify(n, i); for i in range (n): temp = K - B[i]; if (binarySearch( 0 , n - 1 , temp - 1 ) ! = - 1 ): print ( "Found the elements." ); break ; # This code is contributed by Rajput-Ji |
C#
using System; public class GFG { static int []A = { 1, 2, 1, 3, 4 }; static void heapify( int n, int i) { int rootLargest = i; int lchild = 2 * i; int rchild = (2 * i) + 1; if (lchild < n && A[lchild] > A[rootLargest]) rootLargest = lchild; if (rchild < n && A[rchild] > A[rootLargest]) rootLargest = rchild; if (rootLargest != i) { int t = A[i]; A[i] = A[rootLargest]; A[rootLargest] = t; // Recursion heapify(n, rootLargest); } } static int binarySearch( int l, int r, int x) { while (l <= r) { int m = l + (r - l) / 2; if (A[m] == x) return m; if (A[m] < x) l = m + 1; else r = m - 1; } return -1; } public static void Main(String[] args) { int []B = { 3, 1, 5, 1, 2 }; int K = 8; int n = A.Length; // Building the heap for ( int i = n / 2 - 1; i >= 1; i--) heapify(n, i); for ( int i = 0; i < n; i++) // O(n) { int temp = K - B[i]; // O(1) if (binarySearch(0, n - 1, temp - 1) != -1) // O(logn) { Console.Write( "Found the elements." ); break ; } } } } // This code is contributed by Rajput-Ji |
Javascript
<script> function heapify(a,n,i) { let rootLargest = i; let lchild = 2 * i; let rchild = (2 * i) + 1; if (lchild < n && a[lchild] > a[rootLargest]) rootLargest = lchild; if (rchild < n && a[rchild] > a[rootLargest]) rootLargest = rchild; if (rootLargest != i) { swap(a[i] , a[rootLargest]); //Recursion heapify(a , n , rootLargest); } } function binarySearch(a,l,r,x) { while (l <= r) { let m = l + (r - l) / 2; if (a[m] == x) return m; if (a[m] < x) l = m + 1; else r = m - 1; } return -1; } let A = [1,2,1,3,4]; let B = [3,1,5,1,2]; let K = 8; let n = A.length; // Building the heap for (let i = n / 2 - 1 ; i >= 1; i--) heapify(A , n , i); for (let i=0 ; i<n ; i++) //O(n) { let temp = K - B[i]; //O(1) if (binarySearch(A , 0 , n-1 , temp)) //O(logn) { document.write( "Found the elements." ); break ; } } </script> |
输出：
Found the elements
时间复杂度：O（n logn）。
O（对数n+n对数n）
所以，总体时间复杂度=O（n logn）。
一 有效解决方案 这个问题的关键在于 散列 .哈希表是使用 C++中的无序序集 .
- 我们将所有第一个数组元素存储在哈希表中。
- 对于第二个数组的元素，我们从x中减去每个元素，并在哈希表中检查结果。
- 如果结果存在，我们打印元素并输入哈希（第一个数组的元素）。
C++
// C++ program to find all pair in both arrays // whose sum is equal to given value x #include <bits/stdc++.h> using namespace std; // Function to find all pairs in both arrays // whose sum is equal to given value x void findPairs( int arr1[], int arr2[], int n, int m, int x) { // Insert all elements of first array in a hash unordered_set< int > s; for ( int i = 0; i < n; i++) s.insert(arr1[i]); // Subtract sum from second array elements one // by one and check it's present in array first // or not for ( int j = 0; j < m; j++) if (s.find(x - arr2[j]) != s.end()) cout << x - arr2[j] << " " << arr2[j] << endl; } // Driver code int main() { int arr1[] = { 1, 0, -4, 7, 6, 4 }; int arr2[] = { 0, 2, 4, -3, 2, 1 }; int x = 8; int n = sizeof (arr1) / sizeof ( int ); int m = sizeof (arr2) / sizeof ( int ); findPairs(arr1, arr2, n, m, x); return 0; } |
JAVA
// JAVA Code for Given two unsorted arrays, // find all pairs whose sum is x import java.util.*; class GFG { // Function to find all pairs in both arrays // whose sum is equal to given value x public static void findPairs( int arr1[], int arr2[], int n, int m, int x) { // Insert all elements of first array in a hash HashMap<Integer, Integer> s = new HashMap<Integer, Integer>(); for ( int i = 0 ; i < n; i++) s.put(arr1[i], 0 ); // Subtract sum from second array elements one // by one and check it's present in array first // or not for ( int j = 0 ; j < m; j++) if (s.containsKey(x - arr2[j])) System.out.println(x - arr2[j] + " " + arr2[j]); } /* Driver program to test above function */ public static void main(String[] args) { int arr1[] = { 1 , 0 , - 4 , 7 , 6 , 4 }; int arr2[] = { 0 , 2 , 4 , - 3 , 2 , 1 }; int x = 8 ; findPairs(arr1, arr2, arr1.length, arr2.length, x); } } // This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 program to find all # pair in both arrays whose # sum is equal to given value x # Function to find all pairs # in both arrays whose sum is # equal to given value x def findPairs(arr1, arr2, n, m, x): # Insert all elements of # first array in a hash s = set () for i in range ( 0 , n): s.add(arr1[i]) # Subtract sum from second # array elements one by one # and check it's present in # array first or not for j in range ( 0 , m): if ((x - arr2[j]) in s): print ((x - arr2[j]), '', arr2[j]) # Driver code arr1 = [ 1 , 0 , - 4 , 7 , 6 , 4 ] arr2 = [ 0 , 2 , 4 , - 3 , 2 , 1 ] x = 8 n = len (arr1) m = len (arr2) findPairs(arr1, arr2, n, m, x) # This code is contributed # by ihritik |
C#
// C# Code for Given two unsorted arrays, // find all pairs whose sum is x using System; using System.Collections.Generic; class GFG { // Function to find all pairs in // both arrays whose sum is equal // to given value x public static void findPairs( int [] arr1, int [] arr2, int n, int m, int x) { // Insert all elements of first // array in a hash Dictionary< int , int > s = new Dictionary< int , int >(); for ( int i = 0; i < n; i++) { s[arr1[i]] = 0; } // Subtract sum from second array // elements one by one and check // it's present in array first // or not for ( int j = 0; j < m; j++) { if (s.ContainsKey(x - arr2[j])) { Console.WriteLine(x - arr2[j] + " " + arr2[j]); } } } // Driver Code public static void Main( string [] args) { int [] arr1 = new int [] { 1, 0, -4, 7, 6, 4 }; int [] arr2 = new int [] { 0, 2, 4, -3, 2, 1 }; int x = 8; findPairs(arr1, arr2, arr1.Length, arr2.Length, x); } } // This code is contributed by Shrikant13 |
Javascript
<script> // Javascript Code for Given two unsorted arrays, // find all pairs whose sum is x // Function to find all pairs in both arrays // whose sum is equal to given value x function findPairs(arr1, arr2, n, m, x) { // Insert all elements of first array in a hash let s = new Map(); for (let i = 0; i < n; i++) s.set(arr1[i], 0); // Subtract sum from second array elements one // by one and check it's present in array first // or not for (let j = 0; j < m; j++) if (s.has(x - arr2[j])) document.write(x - arr2[j] + " " + arr2[j] + "<br/>" ); } // Driver code let arr1 = [ 1, 0, -4, 7, 6, 4 ]; let arr2 = [ 0, 2, 4, -3, 2, 1 ]; let x = 8; findPairs(arr1, arr2, arr1.length, arr2.length, x); </script> |
输出：
6 24 46 27 1
时间复杂性： O（最大值（n，m）） 辅助空间： O（n）
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