找出消除奇数或替换偶数数组元素的游戏的获胜者

给定一个由N 个整数组成的数组arr[] 。两名玩家，玩家 1和玩家 2 ，逐回合进行游戏，其中一名玩家可以进行以下两种移动之一：

• 将偶数数组元素转换为任何其他整数。
• 删除奇数数组元素。

不能进行任何移动的玩家输掉游戏。任务是打印游戏的获胜者。如果游戏可以永远继续，则打印-1 。

例子：

Input: arr[] = {3, 1, 9, 7}
Output: Player 2
Explanation: Since all array elements are odd, no conversion is possible.
Turn 1: Player 1 deletes 3.
Turn 2: Player 2 deletes 1.
Turn 3: Player 1 deletes 9.
Turn 4: Player 2 deletes 7. Now, Player 1 has no moves left. Therefore, Player 2 wins the game.

Input: arr[]={4, 8}
Output: -1

处理方法：按照以下步骤解决问题：

• 遍历数组。
• 计算数组中存在的偶数和奇数元素的数量。
• 如果偶数元素的数量为零，则执行以下操作：
  • 如果奇数为偶数，则玩家 2将赢得比赛。
  • 否则，玩家 1将赢得比赛。
• 如果奇数为奇数且数组中只有一个偶数元素，则玩家 1将赢得比赛。
• 否则，每次都会有平局。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to evaluate the
// winner of the game
void findWinner(int arr[], int N)
{
    // Stores count of odd
    // array elements
    int odd = 0;
 
    // Stores count of even
    // array elements
    int even = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If arraay element is odd
        if (arr[i] % 2 == 1) {
 
            odd++;
        }
        // Otherwise
        else {
 
            even++;
        }
    }
 
    // If count of even is zero
    if (even == 0) {
 
        // If count of odd is even
        if (odd % 2 == 0) {
 
            cout << "Player 2" << endl;
        }
 
        // If count of odd is odd
        else if (odd % 2 == 1) {
 
            cout << "Player 1" << endl;
        }
    }
 
    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1) {
 
        cout << "Player 1" << endl;
    }
 
    // Otherwise
    else {
 
        cout << -1 << endl;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 1, 9, 7 };
    int N = sizeof(arr)
            / sizeof(arr[0]);
 
    findWinner(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
    
// Function to evaluate the
// winner of the game
static void findWinner(int arr[], int N)
{
     
    // Stores count of odd
    // array elements
    int odd = 0;
 
    // Stores count of even
    // array elements
    int even = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If arraay element is odd
        if (arr[i] % 2 == 1)
        {
            odd++;
        }
         
        // Otherwise
        else
        {
            even++;
        }
    }
 
    // If count of even is zero
    if (even == 0)
    {
         
        // If count of odd is even
        if (odd % 2 == 0)
        {
            System.out.println("Player 2");
        }
 
        // If count of odd is odd
        else if (odd % 2 == 1)
        {
            System.out.println("Player 1");
        }
    }
 
    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1)
    {
        System.out.println("Player 1");
    }
 
    // Otherwise
    else
    {
        System.out.println(-1);
    }
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 3, 1, 9, 7 };
    int N = arr.length;
     
    findWinner(arr, N);
}
}
 
// This code is contributed by ipg2016107

Python3

# Python3 program for the above approach
 
# Function to evaluate the
# winner of the game
def findWinner(arr, N):
     
    # Stores count of odd
    # array elements
    odd = 0
 
    # Stores count of even
    # array elements
    even = 0
 
    # Traverse the array
    for i in range(N):
         
        # If arraay element is odd
        if (arr[i] % 2 == 1):
            odd += 1
 
        # Otherwise
        else:
            even += 1
 
    # If count of even is zero
    if (even == 0):
         
        # If count of odd is even
        if (odd % 2 == 0):
            print("Player 2")
 
        # If count of odd is odd
        elif (odd % 2 == 1):
            print("Player 1")
 
    # If count of odd is odd and
    # count of even is one
    elif (even == 1 and odd % 2 == 1):
        print("Player 1")
 
    # Otherwise
    else:
        print(-1)
 
# Driver code
if __name__ == '__main__':
 
    arr = [ 3, 1, 9, 7 ]
    N = len(arr)
 
    findWinner(arr, N)
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
 
class GFG{
    
// Function to evaluate the
// winner of the game
static void findWinner(int []arr, int N)
{
     
    // Stores count of odd
    // array elements
    int odd = 0;
 
    // Stores count of even
    // array elements
    int even = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If arraay element is odd
        if (arr[i] % 2 == 1)
        {
            odd++;
        }
         
        // Otherwise
        else
        {
            even++;
        }
    }
 
    // If count of even is zero
    if (even == 0)
    {
         
        // If count of odd is even
        if (odd % 2 == 0)
        {
            Console.WriteLine("Player 2");
        }
 
        // If count of odd is odd
        else if (odd % 2 == 1)
        {
            Console.WriteLine("Player 1");
        }
    }
 
    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1)
    {
        Console.WriteLine("Player 1");
    }
 
    // Otherwise
    else
    {
        Console.WriteLine(-1);
    }
}
 
// Driver Code
public static void Main()
{
    int []arr = { 3, 1, 9, 7 };
    int N = arr.Length;
     
    findWinner(arr, N);
}
}
 
// This code is contributed by bgangwar59

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to evaluate the
// winner of the game
function findWinner(arr, N)
{
      
    // Stores count of odd
    // array elements
    let odd = 0;
  
    // Stores count of even
    // array elements
    let even = 0;
  
    // Traverse the array
    for(let i = 0; i < N; i++)
    {
          
        // If arraay element is odd
        if (arr[i] % 2 == 1)
        {
            odd++;
        }
          
        // Otherwise
        else
        {
            even++;
        }
    }
  
    // If count of even is zero
    if (even == 0)
    {
          
        // If count of odd is even
        if (odd % 2 == 0)
        {
            document.write("Player 2");
        }
  
        // If count of odd is odd
        else if (odd % 2 == 1)
        {
            document.write("Player 1");
        }
    }
  
    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1)
    {
        document.write("Player 1");
    }
  
    // Otherwise
    else
    {
        document.write(-1);
    }
}
 
// Driver Code
 
    let arr = [ 3, 1, 9, 7 ];
    let N = arr.length;
      
    findWinner(arr, N);
           

输出：

Player 2

时间复杂度： O(N)
辅助空间：O(1)

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