Question

这是我的参考代码：

import math
def tic_tac_toe():
    board1 = []
    end = False

def choose(): 
   global num
   num = input("How large do you want the grid? (input the SIDE length)")
   for a in range(num**2):
       board1.append(a)

def draw():
    x = num
    v = 0
    for a in range(x):
        print 
        for b in range(x):
            print(board1[v]),
            v = v + 1
    size = num
    rows = [tuple(range(i * size, (i + 1) * size)) for i in range(size)]
    cols = [tuple(range(i, size ** 2 * 1, size)) for i in range(size)]
    diags = [tuple(i * (size + 1) for i in range(size)), tuple(i * (size - 1) for i in range(1, size + 1))]
    global howtowin
    howtowin = tuple(rows + cols + diags)

def p1():
    n = choose_number()
    if board1[n] == 'X' or board1[n] == 'O':
        print("\nAre you stupid... Try again")
        p1()
    else:
        board1[n] = 'X'

def p2():
    n = choose_number()
    if board1[n] == 'X' or board1[n] == 'O':
        print("\nAre you stupid... Try again")
        p2()
    else:
        board1[n] = 'O'

def choose_number():
    while True:
        try:
            a = int(input())
            if a in board1:
                return a
            else:
                print("\nCan you not count to " + str(num**2) + "? Try again")
        except ValueError:
           print("\nI don't speak French! Try again")

def check_board():
    count = 0
    for a in howtowin:
        if board1[a[0]] == board1[a[1]] == board1[a[2]] == 'X':     #check all lengths 3-...
            print("\nPlayer 1 Wins!")
            print("Congratulations!\n")
            return True
        if board1[a[0]] == board1[a[1]] == board1[a[2]] == 'O':
            print("\nPlayer 2 Wins!")
            print("Congratulations!\n")
            return True
    for a in range(num*num):
        if board1[a] == 'X' or board1[a] == 'O':
            count += 1
        if count == num*num:
            print("The game ends in a Tie\n")
            return True

choose()
while not end:
    draw()
    end = check_board()
    if end == True:
        break
    print("\n\nPlayer 1 choose where to place a 'X'")
    p1()
    print
    draw()
    end = check_board()
    if end == True:
        break
    print("\n\nPlayer 2 choose where to place a 'O'")
    p2()
    print

if raw_input("Play again (y/n)\n") == 'y':                                  
    print
    tic_tac_toe()
else:
    print("\nThanks for playing!!!!!!")

tic_tac_toe()

我的问题是check_board（）;我将其硬编码为3的边值。我现在需要它来处理任何边长值。

Answer 1

我发现这是一个非常优雅的解决方案，使用all和any：

def is_winning_row(board, size, player):
    return any((all(board[i][j]==player for j in range(size))) for i in range(size))

如果True player上有board行，False，则此函数将返回size。

从这里，您可能会发现如何使用列，然后使用对角线。

如果您不想将size传递给该函数，只需将len(board)内的board替换为import java.util.Scanner; class binary_diff { public String diff(String st1,String st2) { String nst="",max="";char b='0';boolean tf=(st1.length()>=st2.length()); int l1=st1.length(),l2=st2.length(); if(l1<l2) for(int a=1;a<=l2-l1;a++) st1='0'+st1; else if(l2<l1) for(int a=1;a<=l1-l2;a++) st2="0"+st2; if(!tf)for(int a=l1-1;a>=0;a--) if(st1.charAt(a)!=st2.charAt(a)) if(st2.charAt(a)=='1'){max=st2;st2=st1;st1=max;break;} for(int a=st1.length()-1;a>=0;a--) { if(st1.charAt(a)=='1' && st2.charAt(a)=='0') { if(b=='1') {nst='0'+nst;b='0';} else nst='1'+nst; } else if(st1.charAt(a)==st2.charAt(a) && st2.charAt(a)=='1') { if(b=='1') {nst='1'+nst;b='1';} else nst='0'+nst; } else if(st1.charAt(a)=='0' && st2.charAt(a)=='1') { if(b=='1') nst='0'+nst; else {nst='1'+nst;b='1';} } else { if(b=='1') nst='1'+nst; else nst='0'+nst; } } return nst; } public static void main() { Scanner sc=new Scanner(System.in); System.out.println("Enter the two numbers"); String s1=sc.next(); String s2=sc.next(); binary_diff bd=new binary_diff(); System.out.println(bd.diff(s1,s2)); } }（只要shiva goodshiva goodshiva goodshivagood为{}一个正方形）。

TicTacToe的最终问题

1 个答案: