名人问题 |第 2 组

在 N 人的聚会中，每个人都只认识一个人。这样的人可能出现在聚会上，如果是，（s）他不认识聚会中的任何人。我们只能问“ A知道B吗？ “。在最少的问题中找到陌生人（名人）。
我们可以将问题输入描述为代表聚会中人员的数字/字符数组。我们还有一个假设函数HaveAcquaintance(A, B)如果 A 知道 B，则返回true ，否则返回false 。如何解决问题？

还给出了一个列表know[] ，其中know[i]以{a, b}的形式表示，表示人 a知道人 b 。

例子：

Input: know[] = {{1, 3}, {2, 3}, {4, 3}}, N = 4
Output: 2
Explanation: Person with id = 3 does not know anyone, but every other person knows him.

Input: know[] = {{1, 3}, {2, 3}, {3, 2}, {4, 3}}, N = 4
Output: -1
Explanation: No celebrity is present in the party.

方法：在该问题的Set-1中讨论了朴素的方法和一些优化的方法。

贪心方法：可以使用贪心方法来解决。上面的问题可以通过创建一个bool的向量来解决，遍历给定的矩阵如下：

• 遍历矩阵一次，找到谁都不认识的人。
• 将此存储为当前名人。如果当前名人不是人，则返回-1，否则进行下一步。
• 再次遍历矩阵，现在检查聚会上的每个人是否都知道当前的名人。
• 如果当前名人也符合此标准，则将其作为派对名人返回。
• 否则返回-1。

下面是上述方法的实现

C++

// C++ code to implement above approach
#include 
using namespace std;
 
// Function to find the celebrity
int findPartyCelebrity(int N,
                       vector >& know)
{
 
    vector celebrity(N + 1, false);
    for (int i = 0; i < know.size(); i++)
        celebrity[know[i][0]] = true;
 
    int party_celebrity = -1;
    for (int i = 1; i <= N; i++)
        if (celebrity[i] == false)
            party_celebrity = i;
 
    celebrity.assign(N + 1, false);
    for (int i = 0; i < know.size(); i++)
        if (know[i][1] == party_celebrity)
            celebrity[know[i][0]] = true;
 
    for (int i = 1; i <= N; i++)
        if (i != party_celebrity && celebrity[i] == false)
            party_celebrity = -1;
 
    return party_celebrity;
}
 
// Driver code
int main()
{
    int N = 4;
    vector > know = { { 1, 3 }, { 2, 3 }, { 4, 3 } };
    cout << findPartyCelebrity(N, know);
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG {
 
  // Function to find the celebrity
  static int findPartyCelebrity(int N, int[][] know)
  {
 
    boolean[] celebrity = new boolean[N + 1];
    for (int i = 0; i < know.length; i++)
      celebrity[know[i][0]] = true;
 
    int party_celebrity = -1;
    for (int i = 1; i <= N; i++)
      if (celebrity[i] == false)
        party_celebrity = i;
 
    celebrity = new boolean[N + 1];
    for (int i = 0; i < know.length; i++)
      if (know[i][1] == party_celebrity)
        celebrity[know[i][0]] = true;
 
    for (int i = 1; i <= N; i++)
      if (i != party_celebrity
          && celebrity[i] == false)
        party_celebrity = -1;
 
    return party_celebrity;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 4;
    int[][] know = { { 1, 3 }, { 2, 3 }, { 4, 3 } };
    System.out.println(findPartyCelebrity(N, know));
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python code to implement above approach
 
# Function to find the celebrity
def findPartyCelebrity(N, know):
 
    celebrity = [False] * (N + 1)
    for i in range(len(know)):
        celebrity[know[i][0]] = True
 
    party_celebrity = -1
    for i in range(1, N + 1):
        if celebrity[i] == False:
            party_celebrity = i
 
    celebrity = [False] * (N + 1)
    for i in range(len(know)):
        if know[i][1] == party_celebrity:
            celebrity[know[i][0]] = True
 
    for i in range(1, N + 1):
        if i != party_celebrity and celebrity[i] == False:
            party_celebrity = -1
 
    return party_celebrity
 
# Driver code
N = 4
know = [[1, 3], [2, 3], [4, 3]]
print(findPartyCelebrity(N, know))
 
# This code is contributed by gfgking

C#

// C# code for the above approach
using System;
 
class GFG
{
 
  // Function to find the celebrity
  static int findPartyCelebrity(int N, int[,] know)
  {
 
    bool[] celebrity = new bool[N + 1];
    for (int i = 0; i < know.GetLength(0); i++)
      celebrity[know[i, 0]] = true;
 
    int party_celebrity = -1;
    for (int i = 1; i <= N; i++)
      if (celebrity[i] == false)
        party_celebrity = i;
     
    for(int i = 0; i < N + 1; i++){
        celebrity[i] = false;
    }
     
    for (int i = 0; i < know.GetLength(0); i++)
      if (know[i, 1] == party_celebrity)
        celebrity[know[i, 0]] = true;
 
    for (int i = 1; i <= N; i++)
      if (i != party_celebrity
          && celebrity[i] == false)
        party_celebrity = -1;
 
    return party_celebrity;
  }
 
  // Driver code
  public static void Main()
  {
    int N = 4;
    int[,] know = { { 1, 3 }, { 2, 3 }, { 4, 3 } };
        Console.Write(findPartyCelebrity(N, know));
  }
}
 
// This code is contributed by Samim Hossain Mondal

Javascript

    // JavaScript code to implement above approach
 
    // Function to find the celebrity
    const findPartyCelebrity = (N, know) => {
 
        let celebrity = new Array(N + 1).fill(false);
        for (let i = 0; i < know.length; i++)
            celebrity[know[i][0]] = true;
 
        let party_celebrity = -1;
        for (let i = 1; i <= N; i++)
            if (celebrity[i] == false)
                party_celebrity = i;
 
        celebrity = new Array(N + 1).fill(false);
        for (let i = 0; i < know.length; i++)
            if (know[i][1] == party_celebrity)
                celebrity[know[i][0]] = true;
 
        for (let i = 1; i <= N; i++)
            if (i != party_celebrity && celebrity[i] == false)
                party_celebrity = -1;
 
        return party_celebrity;
    }
 
    // Driver code
    let N = 4;
    let know = [[1, 3], [2, 3], [4, 3]];
    document.write(findPartyCelebrity(N, know));
 
// This code is contributed by rakeshsahni
 

输出：

3

时间复杂度： O(N)
辅助空间： O(N)