给定一个正整数N (代表游戏玩家的数量)和一个字符串arr []数组，该字符串数组由数字字符串组成，这些数字字符串由[‘1’，’N’]范围内的数字组成。考虑给第i个玩家分配了字符串arr [i] ，任务是在所有N个玩家按照以下规则最佳玩游戏时找到游戏的获胜者：
- 玩家1开始游戏，删除字符串arr [1] (基于1的索引)的第一个字符，例如X ，然后在下一回合中，第X位玩家将玩游戏并删除arr [X]的第一个字符，然后很快。
- 无法从分配的字符串删除任何字符的玩家将赢得比赛。
例子：
Input: N = 3, arr[] = { “323”, “2”, “2” }
Output: Player 2
Explanation:
Turn 1: Removing arr[0][0] by Player 1 modifies arr[0] to “23”.
Turn 2: Removing arr[2][0] by Player 3 modifies arr[2] to “”.
Turn 3: Removing arr[1][0] by Player 2 modifies arr[1] to “”.
Turn 4: Player 2 is not able to remove any characters from arr[1].
Therefore, player 2 wins the game.
Input: N = 3, arr[] = { “121”, “21”, “23123” }
Output: Player 1
方法：使用Queue可以有效地删除每个字符串的第一个字符，可以解决该问题。请按照以下步骤解决问题：
- 初始化一个队列数组，例如Q [] ，以使Q [i]存储字符串arr [i]的字符。
- 根据游戏规则使用变量i遍历数组Q []并检查Q [i]中的字符数是否为0 。如果发现是真的，则打印“ Player i” 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the winner of a game of
// repeatedly removing the first character
// to empty a string
void find_Winner(vector& arr, int N)
{
// Store characters of each
// string of the array arr[]
vector > Q(N);
// Stores count of strings in arr[]
int M = arr.size();
// Traverse the array arr[]
for (int i = 0; i < M; i++) {
// Stores length of current string
int len = arr[i].length();
// Traverse the string
for (int j = 0; j < len; j++) {
// Insert arr[i][j]
Q[i].push(arr[i][j] - 1);
}
}
// 1st Player starts the game
int player = 0;
while (Q[player].size() > 0) {
// Stores the player number
// for the next turn
int nextPlayer
= Q[player].front() - '0';
// Remove 1st character of
// current string
Q[player].pop();
// Update player number for
// the next turn
player = nextPlayer;
}
cout << "Player " << (player + 1);
}
// Driver Code
int main()
{
int N = 3;
vector arr
= { "323", "2", "2" };
find_Winner(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find the winner of a game of
// repeatedly removing the first character
// to empty a String
static void find_Winner(String[] arr, int N)
{
// Store characters of each
// String of the array arr[]
@SuppressWarnings("unchecked")
Vector [] Q = new Vector[N];
for (int i = 0; i < Q.length; i++)
Q[i] = new Vector();
// Stores count of Strings in arr[]
int M = arr.length;
// Traverse the array arr[]
for (int i = 0; i < M; i++)
{
// Stores length of current String
int len = arr[i].length();
// Traverse the String
for (int j = 0; j < len; j++)
{
// Insert arr[i][j]
Q[i].add(arr[i].charAt(j));
}
}
// 1st Player starts the game
int player = 0;
while (Q[player].size() > 0 )
{
// Stores the player number
// for the next turn
int nextPlayer
= Q[player].get(0) - '0'-1;
// Remove 1st character of
// current String
Q[player].remove(0);
// Update player number for
// the next turn
player = nextPlayer;
}
System.out.print("Player " + (player + 1));
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
String[] arr
= { "323", "2", "2" };
find_Winner(arr, N);
}
}
// This code is contributed by gauravrajput1 Python3
# Python3 program to implement
# the above approach
# Function to find the winner of a game of
# repeatedly removing the first character
# to empty a string
def find_Winner(arr, N) :
# Store characters of each
# string of the array arr[]
Q = [0]*N
for i in range(N) :
Q[i] = []
# Stores count of strings in arr[]
M = len(arr)
# Traverse the array arr[]
for i in range(M) :
# Stores length of current string
Len = len(arr[i])
# Traverse the string
for j in range(Len) :
# Insert arr[i][j]
Q[i].append(ord(arr[i][j]) - 1)
# 1st Player starts the game
player = 0
while (len(Q[player]) > 0) :
# Stores the player number
# for the next turn
nextPlayer = Q[player][0] - ord('0')
# Remove 1st character of
# current string
del Q[player][0]
# Update player number for
# the next turn
player = nextPlayer
print("Player", (player + 1))
N = 3
arr = [ "323", "2", "2" ]
find_Winner(arr, N)
# This code is contributed by divyeshrabadiya07.C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the winner of a game of
// repeatedly removing the first character
// to empty a String
static void find_Winner(String[] arr, int N)
{
// Store characters of each
// String of the array []arr
List [] Q = new List[N];
for(int i = 0; i < Q.Length; i++)
Q[i] = new List();
// Stores count of Strings in []arr
int M = arr.Length;
// Traverse the array []arr
for(int i = 0; i < M; i++)
{
// Stores length of current String
int len = arr[i].Length;
// Traverse the String
for(int j = 0; j < len; j++)
{
// Insert arr[i,j]
Q[i].Add(arr[i][j]);
}
}
// 1st Player starts the game
int player = 0;
while (Q[player].Count > 0 )
{
// Stores the player number
// for the next turn
int nextPlayer = Q[player][0] - '0'- 1;
// Remove 1st character of
// current String
Q[player].RemoveAt(0);
// Update player number for
// the next turn
player = nextPlayer;
}
Console.Write("Player " + (player + 1));
}
// Driver Code
public static void Main(String[] args)
{
int N = 3;
String[] arr = { "323", "2", "2" };
find_Winner(arr, N);
}
}
// This code is contributed by Rajput-Ji 输出：
Player 2时间复杂度： O(N * M)，其中M是数组中存在的最长字符串的长度。
辅助空间： O(N * M)