通过删除任意两个连续的相似字母找到要赢的游戏的赢家

给定一个由小写字母组成的字符串。

游戏规则：

• 玩家可以选择一对相似的连续字符并将其删除。
• 有两名玩家在玩游戏，最后一步的玩家获胜。

任务是找到获胜者(如果A排名第一且两者都发挥最佳)。

例子：

Input: str = "kaak" 
Output: B
Explanation:
    Initial String: "kaak"
    A's turn:
        removes: "aa"
        Remaining String: "kk"
    B's turn:
        removes: "kk"
        Remaining String: ""
    Since B was the last one to play
    B is the winner.

Input: str = "kk"
Output: A

方法：我们可以使用堆栈来简化问题。

• 每次遇到一个不同于堆栈顶部中存在的字符，都会将其添加到堆栈中。
• 如果堆栈顶部和下一个字符匹配，我们将从堆栈中弹出字符并增加计数。
• 最后，我们只需要通过检查count％2来查看谁获胜。

下面是上述方法的实现：

C++

#include 
using namespace std;
  
// Function to play the game 
// and find the winner
void findWinner(string s)
{
    int i, count = 0, n;
    n = s.length();
    stack st;
  
    // ckecking the top of the stack with
    // the i th character of the string
    // add it to the stack if they are different
    // otherwise increment count
    for (i = 0; i < n; i++) {
        if (st.empty() || st.top() != s[i]) {
            st.push(s[i]);
        }
        else {
            count++;
            st.pop();
        }
    }
  
    // Check who has won
    if (count % 2 == 0) {
        cout << "B" << endl;
    }
    else {
        cout << "A" << endl;
    }
}
  
// Driver code
int main()
{
    string s = "kaak";
  
    findWinner(s);
  
    return 0;
}

Java

// Java implementation for above approach
import java.util.*;
  
class GFG 
{
  
// Function to play the game 
// and find the winner
static void findWinner(String s) 
{
    int i, count = 0, n;
    n = s.length();
    Stack st = new Stack();
  
    // ckecking the top of the stack with
    // the i th character of the string
    // add it to the stack if they are different
    // otherwise increment count
    for (i = 0; i < n; i++) 
    {
        if (st.isEmpty() || 
            st.peek() != s.charAt(i))
        {
            st.push(s.charAt(i));
        } 
        else
        {
            count++;
            st.pop();
        }
    }
  
    // Check who has won
    if (count % 2 == 0) 
    {
        System.out.println("B");
    } 
    else
    {
        System.out.println("A");
    }
}
  
// Driver code
public static void main(String[] args) 
{
    String s = "kaak";
  
    findWinner(s);
}
} 
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
  
# Function to play the game 
# and find the winner 
def findWinner(s) : 
  
    count = 0
    n = len(s); 
    st = []; 
  
    # ckecking the top of the stack with 
    # the i th character of the string 
    # add it to the stack if they are different 
    # otherwise increment count 
    for i in range(n) :
        if (len(st) == 0 or st[-1] != s[i]) : 
            st.append(s[i]); 
              
        else : 
            count += 1; 
            st.pop(); 
  
    # Check who has won 
    if (count % 2 == 0) :
        print("B"); 
      
    else :
        print("A"); 
          
# Driver code 
if __name__ == "__main__" : 
  
    s = "kaak"; 
  
    findWinner(s);
  
# This code is contributed by AnkitRai01

C#

// C# implementation for above approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
// Function to play the game 
// and find the winner
static void findWinner(String s) 
{
    int i, count = 0, n;
    n = s.Length;
    Stack st = new Stack();
  
    // ckecking the top of the stack with
    // the i th character of the string
    // add it to the stack if they are different
    // otherwise increment count
    for (i = 0; i < n; i++) 
    {
        if (st.Count == 0 || 
            st.Peek() != s[i])
        {
            st.Push(s[i]);
        } 
        else
        {
            count++;
            st.Pop();
        }
    }
  
    // Check who has won
    if (count % 2 == 0) 
    {
        Console.WriteLine("B");
    } 
    else
    {
        Console.WriteLine("A");
    }
}
  
// Driver code
public static void Main(String[] args) 
{
    String s = "kaak";
  
    findWinner(s);
}
} 
  
// This code is contributed by 29AjayKumar

输出：

B