用负数、正数和 0 个产品制作三个非空集合

你得到一个由 n 个不同整数组成的数组。你的任务是将这个数组分成三个非空集，以便满足以下条件：

1. 第一个集合应包含其乘积应小于 0 的元素。
2. 第二组应包含其乘积应大于 0 的元素。
3. 第三组应包含其乘积应等于 0 的元素。

笔记：-

1. 在给定数组中，每个数字必须恰好出现在一组中。
2. 可以假设我们总是可以制作三个集合（输入数组中至少有一个负元素和一个 0）。

例子：

Input : 4
        arr[] = -1 -2 -3 0
Output :  -1
          -3 -2 
           0
In this example, product of first
set is negative, product of second
set is positive and product of third
set is 0.

在这个问题中，我们只需要将输入数据拆分为 3 个向量：第一个将包含负数，第二个正数，第三个零。如果第一个向量的大小是偶数，则将一个数字从它移到第三个向量。如果第二个向量仅包含 1，则将两个数字从第一个向量移动到第二个向量。

C++

// CPP program to make three non-empty sets
// as per the given conditions.
#include 
using namespace std;
 
void makeSets(int arr[], int n)
{
    vector first, second, third;
 
    // insert number equal to 0 to third set.
    // numbers greater than 0 to second set.
    // insert numbers less than 0 to first set.
    for (int i = 0; i < n; i++) {
        if (arr[i] == 0)
            third.push_back(arr[i]);
        if (arr[i] > 0)
            second.push_back(arr[i]);
        if (arr[i] < 0)
            first.push_back(arr[i]);
    }
 
    if (first.size() == 0 || third.size() == 0)
    {
        cout << "Not Possible";
        return;
    }
 
    // if second set is empty.
    if (second.size() == 0) {
        for (int i = 0; i < 2; i++)
        {
            second.push_back(first.back());
            first.pop_back();
        }
    }
 
    // if length of first set is even.
    if (first.size() % 2 == 0) {
        third.push_back(first.back());
        first.pop_back();
    }
 
    // output the first set elements.
    for (int i = 0; i < first.size(); i++)
        cout << first[i] << " ";   
     
    // output the second set elements.
    cout << endl;
    for (int i = 0; i < second.size(); i++)
        cout << second[i] << " ";   
 
    // output the third set elements.
    cout << endl;
    for (int i = 0; i < third.size(); i++)
        cout << third[i] << " ";   
}
 
// Driver Function
int main()
{
    int arr[] = { -1, 2, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    makeSets(arr, n);
    return 0;
}

Java

// Java program to make three non-empty sets
// as per the given condition.
import java.util.*;
 
class GFG
{
 
    static void makeSets(int arr[], int n)
    {
        Vector first = new Vector();
        Vector second = new Vector();
        Vector third = new Vector();
 
        // insert number equal to 0 to third set.
        // numbers greater than 0 to second set.
        // insert numbers less than 0 to first set.
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == 0)
            {
                third.add(arr[i]);
            }
            if (arr[i] > 0)
            {
                second.add(arr[i]);
            }
            if (arr[i] < 0)
            {
                first.add(arr[i]);
            }
        }
 
        if (first.size() == 0 || third.size() == 0)
        {
            System.out.print("Not Possible");
            return;
        }
 
        // if second set is empty.
        if (second.size() == 0)
        {
            for (int i = 0; i < 2; i++)
            {
                second.add(first.lastElement());
                first.remove(first.size() - 1);
            }
        }
 
        // if length of first set is even.
        if (first.size() % 2 == 0)
        {
            third.add(first.lastElement());
            first.remove(first.size() - 1);
        }
 
        // output the first set elements.
        for (int i = 0; i < first.size(); i++)
        {
            System.out.print(first.get(i) + " ");
        }
 
        // output the second set elements.
        System.out.println();
        for (int i = 0; i < second.size(); i++)
        {
            System.out.print(second.get(i) + " ");
        }
 
        // output the third set elements.
        System.out.println();
        for (int i = 0; i < third.size(); i++)
        {
            System.out.print(third.get(i) + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {-1, 2, 0};
        int n = arr.length;
        makeSets(arr, n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to make three non-empty sets
# as per the given conditions.
def makeSets(arr, n):
    first = []
    second = []
    third = []
     
    # insert number equal to 0 to third set.
    # numbers greater than 0 to second set.
    # insert numbers less than 0 to first set.
    for i in range(n):
        if (arr[i] == 0):
            third.append(arr[i])
        if (arr[i] > 0):
            second.append(arr[i])
        if (arr[i] < 0):
            first.append(arr[i])
     
    if (len(first) == 0 or len(third) == 0):
        print("Not Possible")
        return
     
    # if second set is empty.
    if (len(second)== 0):
        for i in range(2):
            second.append(first[-1])
            first.pop()
             
    # if length of first set is even.
    if (len(first) % 2 == 0):
        third.append(first[-1])
        first.pop()
     
    # output the first set elements.
    for i in range(len(first)):
        print(first[i], end = " ")
     
    # output the second set elements.
    print()
    for i in range(len(second)):
        print(second[i], end = " ")
         
    # output the third set elements.
    print()
    for i in range(len(third)):
        print(third[i], end = " ")
         
# Driver Function
arr = [-1, 2, 0]
n = len(arr)
makeSets(arr, n)
 
# This Code is contributed by SHUBHAMSINGH10

C#

// C# program to make three non-empty sets
// as per the given condition.
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static void makeSets(int []arr, int n)
    {
        List first = new List();
        List second = new List();
        List third = new List();
 
        // insert number equal to 0 to third set.
        // numbers greater than 0 to second set.
        // insert numbers less than 0 to first set.
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == 0)
            {
                third.Add(arr[i]);
            }
            if (arr[i] > 0)
            {
                second.Add(arr[i]);
            }
            if (arr[i] < 0)
            {
                first.Add(arr[i]);
            }
        }
 
        if (first.Count == 0 || third.Count == 0)
        {
            Console.Write("Not Possible");
            return;
        }
 
        // if second set is empty.
        if (second.Count == 0)
        {
            for (int i = 0; i < 2; i++)
            {
                second.Add(first[first.Count-1]);
                first.Remove(first.Count - 1);
            }
        }
 
        // if length of first set is even.
        if (first.Count % 2 == 0)
        {
            third.Add(first[first.Count-1]);
            first.Remove(first.Count - 1);
        }
 
        // output the first set elements.
        for (int i = 0; i < first.Count; i++)
        {
            Console.Write(first[i] + " ");
        }
 
        // output the second set elements.
        Console.WriteLine();
        for (int i = 0; i < second.Count; i++)
        {
            Console.Write(second[i] + " ");
        }
 
        // output the third set elements.
    Console.WriteLine();
        for (int i = 0; i < third.Count; i++)
        {
            Console.Write(third[i] + " ");
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {-1, 2, 0};
        int n = arr.Length;
        makeSets(arr, n);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

 
// JavaScript program to make three non-empty sets
// as per the given conditions.
 
 
function makeSets(arr, n) {
    let first = [], second = [], third = [];
 
    // insert number equal to 0 to third set.
    // numbers greater than 0 to second set.
    // insert numbers less than 0 to first set.
    for (let i = 0; i < n; i++) {
        if (arr[i] == 0)
            third.push(arr[i]);
        if (arr[i] > 0)
            second.push(arr[i]);
        if (arr[i] < 0)
            first.push(arr[i]);
    }
 
    if (first.length == 0 || third.length == 0) {
        document.write("Not Possible");
        return;
    }
 
    // if second set is empty.
    if (second.length == 0) {
        for (let i = 0; i < 2; i++) {
            second.push(first[first.length - 1]);
            first.pop();
        }
    }
 
    // if length of first set is even.
    if (first.length % 2 == 0) {
        third.push(first[first.length - 1]);
        first.pop();
    }
 
    // output the first set elements.
    for (let i = 0; i < first.length; i++)
        document.write(first[i] + " ");
 
    // output the second set elements.
    document.write("<br>");
    for (let i = 0; i < second.length; i++)
        document.write(second[i] + " ");
 
    // output the third set elements.
    document.write("<br>");
    for (let i = 0; i < third.length; i++)
        document.write(third[i] + " ");
}
 
// Driver Function
 
let arr = [-1, 2, 0];
let n = arr.length;
makeSets(arr, n);
 

输出：

-1 
2 
0

时间复杂度：- O(n)