给定一个由二进制字符串组成的数组arr[] ,任务是在两个玩家按照以下规则进行最佳游戏时找到游戏的获胜者:
- 玩家 1开始游戏。
- 在每一轮中,玩家必须选择一个非空字符串并从字符串的开头删除正数个字符。
- 玩家 1只能选择以字符’0′ 开头的字符串,而玩家 2只能选择以字符’1′ 开头的字符串。
- 不能移动的玩家输掉游戏。
例子:
Input: arr[] = {“010”, “101”}
Output: Player 2
Explanation:
First move for player 1 = {0, 101}
First move for player 2 = {0, 1}
Second move for player 1 = {1}
Second move for player 2 = {}
No moves left for player 1.
Therefore player2 wins.
Input: arr[] = {“010”, “001”}
Output: Player 1
方法:这个想法是比较如果两个玩家都以最佳方式玩游戏,每个玩家可以进行的移动总数。请按照以下步骤操作:
- 如果在任何字符串中连续出现相同的字符,那么只需用该字符的单个出现替换它们,因为最好删除开头出现的所有字符。
- 现在,如果字符串的起始元素与其最后一个元素相同,那么即使没有这个字符串,游戏的场景也保持不变,因为如果一个玩家在这个字符串上移动,另一个玩家通过删除字符进行下一步来自相同的字符串,导致第一个玩家的位置完全相同。
- 如果字符串的起始元素与其最后一个元素不同,则需要玩家多走一步。
- 因此,只需计算每个玩家必须进行的额外移动次数即可。
- 用完额外动作的玩家将输掉比赛。
下面是上述方法的实现:
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to find the player who
// loses the game
void findPlayer(string str[], int n)
{
// Moves for the first player
int move_first = 0;
// Moves for the second player
int move_sec = 0;
// Iterate over array of strings
for (int i = 0; i < n; i++) {
// Check if the first and last
// character are the same
if (str[i][0]
== str[i][str[i].length() - 1]) {
// Check if string start and
// end with character '0'
if (str[i][0] == 48)
move_first++;
else
move_sec++;
}
}
// If first player have less moves
if (move_first <= move_sec) {
cout << "Player 2 wins";
}
else {
cout << "Player 1 wins";
}
}
// Driver Code
int main()
{
// Given array of strings
string str[] = { "010", "101" };
int N = sizeof(str)
/ sizeof(str[0]);
// Function Call
findPlayer(str, N);
return 0;
}
Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to find the player who
// loses the game
static void findPlayer(String str[],
int n)
{
// Moves for the
// first player
int move_first = 0;
// Moves for the
// second player
int move_sec = 0;
// Iterate over array
// of Strings
for (int i = 0; i < n - 1; i++)
{
// Check if the first and last
// character are the same
if (str[i].charAt(0) ==
str[i].charAt(str[i].length() - 1))
{
// Check if String start and
// end with character '0'
if (str[i].charAt(0) == 48)
move_first++;
else
move_sec++;
}
}
// If first player have less moves
if (move_first <= move_sec)
{
System.out.print("Player 2 wins");
}
else
{
System.out.print("Player 1 wins");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array of Strings
String str[] = {"010", "101"};
int N = str[0].length();
// Function Call
findPlayer(str, N);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program for the above approach
# Function to find the player who
# loses the game
def findPlayer(str, n):
# Moves for the first player
move_first = 0
# Moves for the second player
move_sec = 0
# Iterate over array of strings
for i in range(n):
# Check if the first and last
# character are the same
if (str[i][0] ==
str[i][len(str[i]) - 1]):
# Check if string start and
# end with character '0'
if (str[i][0] == 48):
move_first += 1
else:
move_sec += 1
# If first player have less moves
if (move_first <= move_sec):
print("Player 2 wins")
else:
print("Player 1 wins")
# Driver Code
# Given array of strings
str = [ "010", "101" ]
N = len(str)
# Function call
findPlayer(str, N)
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the player who
// loses the game
static void findPlayer(string[] str, int n)
{
// Moves for the first player
int move_first = 0;
// Moves for the second player
int move_sec = 0;
// Iterate over array of strings
for(int i = 0; i < n; i++)
{
// Check if the first and last
// character are the same
if (str[i][0] ==
str[i][str[i].Length - 1])
{
// Check if string start and
// end with character '0'
if ((str[i][0]) == 48)
move_first++;
else
move_sec++;
}
}
// If first player have less moves
if (move_first <= move_sec)
{
Console.Write("Player 2 wins");
}
else
{
Console.Write("Player 1 wins");
}
}
// Driver Code
public static void Main ()
{
// Given array of strings
string[] str = { "010", "101" };
int N = str.Length;
// Function call
findPlayer(str, N);
}
}
// This code is contributed by sanjoy_62
Javascript
输出:
Player 2 wins
时间复杂度: O(N)
辅助空间: O(1)
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