找出三个数组中至少两个数组中的数字
原文:https://www . geesforgeks . org/find-numbers-在三个数组中的至少两个数组中出现/
给定 3 个数组、arr【】、brr【】和crr【】，任务是从给定的 3 个数组 示例中找到至少 2 个数组中的公共元素:
输入 : arr[] = {1，1，3，2，4}，brr[] = {2，3，5}，crr[] = {3，6} 输出 : {2，3} 解释:元素 2 和 3 出现在至少 2 个数组中
输入:【arr[]= { 3.1}，【br[]= { 2.3 }，【crr[]= { 1.2 } 输出【2，3，1 })****
**接近:可以借助 HashSet 解决任务，按照以下步骤解决问题:****
- *创建三个 hashsets、s2 和 s3)来存储三个数组的不同元素，还创建一个 HashSet 来存储所有三个数组的不同元素。*
- *迭代集合，检查至少两个集合(s1，s2)、(s1，s3)和(s2，s3)中的公共元素。*
- *如果一个元素满足上述条件，打印该元素。*
*下面是上述代码的实现:*
*C++*
**// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
void get(vector<int>& p, vector<int>& c, vector<int>& m)
{
// Store distinct elements of array arr[]
set<int> s1;
// Store distinct elements of array brr[]
set<int> s2;
// Store distinct elements of array crr[]
set<int> s3;
// Store the distinct elements
// of all three arrays
set<int> set;
for (auto i : p) {
s1.insert(i);
set.insert(i);
}
for (auto i : c) {
s2.insert(i);
set.insert(i);
}
for (int i : m) {
s3.insert(i);
set.insert(i);
}
// Stores the common elements
vector<int> result;
for (auto i : set) {
if (s1.count(i) && s2.count(i)
|| s2.count(i) && s3.count(i)
|| s1.count(i) && s3.count(i))
cout << i << " ";
}
}
// Driver Code
int main()
{
vector<int> arr = { 1, 1, 3, 2, 4 };
vector<int> brr = { 2, 3, 5 };
vector<int> crr = { 3, 6 };
get(arr, brr, crr);
return 0;
}
// This code is contributed by rakeshsahni**
*Java 语言(一种计算机语言，尤用于创建网站)*
**// Java implementation of above approach
import java.io.*;
import java.util.*;
class GFG {
public static void get(
int[] p, int[] c, int[] m)
{
// Store distinct elements of array arr[]
Set<Integer> s1 = new HashSet<>();
// Store distinct elements of array brr[]
Set<Integer> s2 = new HashSet<>();
// Store distinct elements of array crr[]
Set<Integer> s3 = new HashSet<>();
// Store the distinct elements
// of all three arrays
Set<Integer> set = new HashSet<>();
for (int i : p) {
s1.add(i);
set.add(i);
}
for (int i : c) {
s2.add(i);
set.add(i);
}
for (int i : m) {
s3.add(i);
set.add(i);
}
// Stores the common elements
List<Integer> result
= new ArrayList<>();
for (int i : set) {
if (s1.contains(i)
&& s2.contains(i)
|| s2.contains(i)
&& s3.contains(i)
|| s1.contains(i)
&& s3.contains(i))
System.out.print(i + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 1, 3, 2, 4 };
int[] brr = { 2, 3, 5 };
int[] crr = { 3, 6 };
get(arr, brr, crr);
}
}**
*Python 3*
**# Python3 implementation of above approach
def get(p, c, m) :
# Store distinct elements of array arr[]
s1 = set();
# Store distinct elements of array brr[]
s2 = set();
# Store distinct elements of array crr[]
s3 = set();
# Store the distinct elements
# of all three arrays
set_obj = set();
for i in p :
s1.add(i);
set_obj.add(i);
for i in c :
s2.add(i);
set_obj.add(i);
for i in m :
s3.add(i);
set_obj.add(i);
# Stores the common elements
result = [];
for i in set_obj :
if (list(s1).count(i) and list(s2).count(i)
or list(s2).count(i) and list(s3).count(i)
or list(s1).count(i) and list(s3).count(i)) :
print(i,end= " ");
# Driver Code
if __name__ == "__main__" :
arr = [ 1, 1, 3, 2, 4 ];
brr = [ 2, 3, 5 ];
crr = [ 3, 6 ];
get(arr, brr, crr);
# This code is contributed by AnkThon**
*C#*
**// C# implementation of above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG {
static void get(
int[] p, int[] c, int[] m)
{
// Store distinct elements of array arr[]
HashSet<int> s1 = new HashSet<int>();
// Store distinct elements of array brr[]
HashSet<int> s2 = new HashSet<int>();
// Store distinct elements of array crr[]
HashSet<int> s3 = new HashSet<int>();
// Store the distinct elements
// of all three arrays
HashSet<int> set = new HashSet<int>();
foreach (int i in p) {
s1.Add(i);
set.Add(i);
}
foreach (int i in c) {
s2.Add(i);
set.Add(i);
}
foreach (int i in m) {
s3.Add(i);
set.Add(i);
}
// Stores the common elements
ArrayList result
= new ArrayList();
foreach (int i in set) {
if (s1.Contains(i)
&& s2.Contains(i)
|| s2.Contains(i)
&& s3.Contains(i)
|| s1.Contains(i)
&& s3.Contains(i))
Console.Write(i + " ");
}
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 1, 3, 2, 4 };
int[] brr = { 2, 3, 5 };
int[] crr = { 3, 6 };
get(arr, brr, crr);
}
}
// This code is contributed by Samim Hossain Mondal.**
*java 描述语言*
**<script>
// Javascript implementation of above approach
function get(p, c, m)
{
// Store distinct elements of array arr[]
let s1 = new Set();
// Store distinct elements of array brr[]
let s2 = new Set();
// Store distinct elements of array crr[]
let s3 = new Set();
// Store the distinct elements
// of all three arrays
let set = new Set();
for (i of p) {
s1.add(i);
set.add(i);
}
for (i of c) {
s2.add(i);
set.add(i);
}
for (i of m) {
s3.add(i);
set.add(i);
}
// Stores the common elements
let result = [];
for (i of [...set].reverse()) {
if (s1.has(i) && s2.has(i)
|| s2.has(i) && s3.has(i)
|| s1.has(i) && s3.has(i))
document.write(i + " ");
}
}
// Driver Code
let arr = [1, 1, 3, 2, 4];
let brr = [2, 3, 5];
let crr = [3, 6];
get(arr, brr, crr);
// This code is contributed by gfgking
</script>**
**Output
2 3****
*时间复杂度 : O(n1+n2+n3)(其中 n1、n2 和 n3 为给定数组的长度) 辅助空间 : O(n1+n2+n3)*
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