找到最后一个能够翻转二进制字符串中的字符的玩家

给定一个长度为N的二进制字符串S ，如果两个玩家A和B按照以下规则进行最佳游戏，则任务是找到游戏的获胜者：

• 玩家 A 总是开始游戏。
• 在玩家的第一个回合中，他可以移动到由‘0’组成的任何索引(基于 1 的索引)并使其成为‘1’ 。
• 对于随后的回合，如果任何玩家处于索引i ，那么他可以移动到它的相邻索引之一，如果它包含0 ，并在移动后将其转换为‘1’ 。
• 如果任何玩家在他的回合中无法移动到任何位置，则该玩家输掉游戏。

任务是找到游戏的赢家。

例子：

Input: S = “1100011”
Output: Player A
Explanation:
The indices 3, 4 and 5 consists of 0s and indices 1, 2, 6 and 7 consists of 1s. 
A starts by flipping the character at index 4..
B flips either the index 3 or 5.
A is now left with only one index adjacent to 4, which B did not pick. After A flips the character at that index, B does not have any character to flip. Since B has no moves, A wins.
Hence, print “Player A”.

Input: S = “11111”
Output: Player B

方法：这个想法是将给定数组arr[]中仅由0组成的所有子字符串的长度存储在另一个数组中，比如V[] 。现在，出现以下情况：

1. 如果 V 的大小为 0 ：在这种情况下，数组不包含任何0 。因此，玩家 A不能采取任何行动并输掉比赛。因此，打印Player B 。
2. 如果 V 的大小为 1 ：在这种情况下，有 1 个仅由0组成的子字符串，例如长度为L 。如果L 的值为奇数，则玩家 A赢得比赛。否则，玩家 B赢得比赛。
3. 在所有其他情况下：分别在first和second中存储0 s 的最大和第二大连续段的长度。当且仅当first 的值为奇数且(first + 1)/2 > second 时，玩家 A才能赢得比赛。否则，玩家 B赢得比赛。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if player A wins
// the game or not
void findWinner(string a, int n)
{
    // Stores size of the groups of 0s
    vector v;
 
    // Stores size of the group of 0s
    int c = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // Increment c by 1 if a[i] is 0
        if (a[i] == '0') {
            c++;
        }
 
        // Otherwise, push the size
        // in array and reset c to 0
        else {
            if (c != 0)
                v.push_back(c);
            c = 0;
        }
    }
    if (c != 0)
        v.push_back(c);
 
    // If there is no substring of
    // odd length consisting only of 0s
    if (v.size() == 0) {
        cout << "Player B";
        return;
    }
 
    // If there is only 1 substring of
    // odd length consisting only of 0s
    if (v.size() == 1) {
        if (v[0] & 1)
            cout << "Player A";
 
        // Otherwise
        else
            cout << "Player B";
        return;
    }
 
    // Stores the size of the largest
    // and second largest substrings of 0s
    int first = INT_MIN;
    int second = INT_MIN;
 
    // Traverse the array v[]
    for (int i = 0; i < v.size(); i++) {
 
        // If current element is greater
        // than first, then update both
        // first and second
        if (a[i] > first) {
            second = first;
            first = a[i];
        }
 
        // If arr[i] is in between
        // first and second, then
        // update second
        else if (a[i] > second
                 && a[i] != first)
            second = a[i];
    }
 
    // If the condition is satisfied
    if ((first & 1)
        && (first + 1) / 2 > second)
        cout << "Player A";
    else
        cout << "Player B";
}
 
// Driver Code
int main()
{
    string S = "1100011";
    int N = S.length();
    findWinner(S, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to check if player A wins
  // the game or not
  static void findWinner(String a, int n)
  {
 
    // Stores size of the groups of 0s
    Vector v = new Vector(); 
 
    // Stores size of the group of 0s
    int c = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++)
    {
 
      // Increment c by 1 if a[i] is 0
      if (a.charAt(i) == '0')
      {
        c++;
      }
 
      // Otherwise, push the size
      // in array and reset c to 0
      else
      {
        if (c != 0)
          v.add(c);
        c = 0;
      }
    }
    if (c != 0)
      v.add(c);
 
    // If there is no substring of
    // odd length consisting only of 0s
    if (v.size() == 0)
    {
      System.out.print("Player B");
      return;
    }
 
    // If there is only 1 substring of
    // odd length consisting only of 0s
    if (v.size() == 1)
    {
      if ((v.get(0) & 1) != 0)
        System.out.print("Player A");
 
      // Otherwise
      else
        System.out.print("Player B");
      return;
    }
 
    // Stores the size of the largest
    // and second largest substrings of 0s
    int first = Integer.MIN_VALUE;
    int second = Integer.MIN_VALUE;
 
    // Traverse the array v[]
    for (int i = 0; i < v.size(); i++)
    {
 
      // If current element is greater
      // than first, then update both
      // first and second
      if (a.charAt(i) > first) {
        second = first;
        first = a.charAt(i);
      }
 
      // If arr[i] is in between
      // first and second, then
      // update second
      else if (a.charAt(i) > second
               && a.charAt(i) != first)
        second = a.charAt(i);
    }
 
    // If the condition is satisfied
    if ((first & 1) != 0
        && (first + 1) / 2 > second)
      System.out.print("Player A");
    else
      System.out.print("Player B");
  }
 
  // Driver code
  public static void main(String[] args)
  {
    String S = "1100011";
    int N = S.length();
    findWinner(S, N);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Python3

# Python3 program for the above approach
import sys
 
# Function to check if player A wins
# the game or not
def findWinner(a, n) :
 
    # Stores size of the groups of 0s
    v = []
 
    # Stores size of the group of 0s
    c = 0
 
    # Traverse the array
    for i in range(0, n) :
 
        # Increment c by 1 if a[i] is 0
        if (a[i] == '0') :
            c += 1
 
        # Otherwise, push the size
        # in array and reset c to 0
        else :
            if (c != 0) :
                v.append(c)
            c = 0
     
    if (c != 0) :
        v.append(c)
 
    # If there is no substring of
    # odd length consisting only of 0s
    if (len(v) == 0) :
        print("Player B", end = "")
        return
 
    # If there is only 1 substring of
    # odd length consisting only of 0s
    if (len(v) == 1) :
        if ((v[0] & 1) != 0) :
            print("Player A", end = "")
 
        # Otherwise
        else :
            print("Player B", end = "")
        return
 
    # Stores the size of the largest
    # and second largest substrings of 0s
    first = sys.minsize
    second = sys.minsize
 
    # Traverse the array v[]
    for i in range(len(v)) :
 
        # If current element is greater
        # than first, then update both
        # first and second
        if (a[i] > first) :
            second = first
            first = a[i]
 
        # If arr[i] is in between
        # first and second, then
        # update second
        elif (a[i] > second and a[i] != first) :
            second = a[i]
 
    # If the condition is satisfied
    if (((first & 1) != 0) and (first + 1) // 2 > second) :
        print("Player A", end = "")
    else :
        print("Player B", end = "")
 
S = "1100011"
N = len(S)
findWinner(S, N)
 
# This code is contributed by divyesh072019.

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG{
 
  // Function to check if player A wins
  // the game or not
  static void findWinner(string a, int n)
  {
 
    // Stores size of the groups of 0s
    List v = new List(); 
 
    // Stores size of the group of 0s
    int c = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++)
    {
 
      // Increment c by 1 if a[i] is 0
      if (a[i] == '0')
      {
        c++;
      }
 
      // Otherwise, push the size
      // in array and reset c to 0
      else
      {
        if (c != 0)
          v.Add(c);
        c = 0;
      }
    }
    if (c != 0)
      v.Add(c);
 
    // If there is no substring of
    // odd length consisting only of 0s
    if (v.Count == 0)
    {
      Console.Write("Player B");
      return;
    }
 
    // If there is only 1 substring of
    // odd length consisting only of 0s
    if (v.Count == 1)
    {
      if ((v[0] & 1) != 0)
        Console.Write("Player A");
 
      // Otherwise
      else
        Console.Write("Player B");
      return;
    }
 
    // Stores the size of the largest
    // and second largest substrings of 0s
    int first = Int32.MinValue;
    int second = Int32.MinValue;
 
    // Traverse the array v[]
    for (int i = 0; i < v.Count; i++)
    {
 
      // If current element is greater
      // than first, then update both
      // first and second
      if (a[i] > first) {
        second = first;
        first = a[i];
      }
 
      // If arr[i] is in between
      // first and second, then
      // update second
      else if (a[i] > second
               && a[i] != first)
        second = a[i];
    }
 
    // If the condition is satisfied
    if ((first & 1) != 0
        && (first + 1) / 2 > second)
      Console.Write("Player A");
    else
      Console.Write("Player B");
  }
 
 
// Driver Code
public static void Main(String[] args)
{
    string S = "1100011";
    int N = S.Length;
    findWinner(S, N);
}
}
 
// This code is contributed by splevel62.

Javascript

    // Javascript program to implement the above approach
     
    // Function to check if player A wins
    // the game or not
    function findWinner(a, n)
    {
 
      // Stores size of the groups of 0s
      let v = [];
 
      // Stores size of the group of 0s
      let c = 0;
 
      // Traverse the array
      for (let i = 0; i < n; i++)
      {
 
        // Increment c by 1 if a[i] is 0
        if (a[i] == '0')
        {
          c++;
        }
 
        // Otherwise, push the size
        // in array and reset c to 0
        else
        {
          if (c != 0)
            v.push(c);
          c = 0;
        }
      }
      if (c != 0)
        v.push(c);
 
      // If there is no substring of
      // odd length consisting only of 0s
      if (v.length == 0)
      {
        document.write("Player B");
        return;
      }
 
      // If there is only 1 substring of
      // odd length consisting only of 0s
      if (v.length == 1)
      {
        if ((v[0] & 1) != 0)
          document.write("Player A");
 
        // Otherwise
        else
          document.write("Player B");
        return;
      }
 
      // Stores the size of the largest
      // and second largest substrings of 0s
      let first = Number.MIN_VALUE;
      let second = Number.MIN_VALUE;
 
      // Traverse the array v[]
      for (let i = 0; i < v.length; i++)
      {
 
        // If current element is greater
        // than first, then update both
        // first and second
        if (a[i] > first) {
          second = first;
          first = a[i];
        }
 
        // If arr[i] is in between
        // first and second, then
        // update second
        else if (a[i] > second && a[i] != first)
          second = a[i];
      }
 
      // If the condition is satisfied
      if ((first & 1) != 0 && parseInt((first + 1) / 2, 10) > second)
        document.write("Player A");
      else
        document.write("Player B");
    }
     
    let S = "1100011";
    let N = S.length;
    findWinner(S, N);
   
  // This code is contributed by suresh07.

输出：

Player A

时间复杂度： O(N)
辅助空间： O(N)

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