python实现2048小游戏

2048的python/' target='_blank'>python实现。修改自某网友的代码，解决了原网友版本的两个小bug：

1. 原版游戏每次只消除一次，而不是递归消除。如 [2 ,2 ,2 ,2] 左移动的话应该是 [4, 4, 0, 0] , 而不是[8 , 0 , 0 ,0]
2. 对游戏结束的侦测有bug，已经改正。

2048game.py

# -*- coding: utf-8 -*-"""Created on Tue Jul 1 14:15:39 2014 @author: kelvin""" import random class game2048: totalScore = 0 v = [[2, 8, 8, 2],   [4, 2, 4, 8],   [2, 4, 2, 0],   [4, 2, 4, 0]] ''' v = [[0, 0, 0, 0],   [0, 0, 0, 0],   [0, 0, 0, 0],   [0, 0, 0, 0]] ''' def __init__(self):  for i in range(4):   self.v[i] = [random.choice([0,0,0,2,2,4]) for x in range(4)]   def display(self):  print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[0][0], self.v[0][1], self.v[0][2], self.v[0][3]))  print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[1][0], self.v[1][1], self.v[1][2], self.v[1][3]))  print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[2][0], self.v[2][1], self.v[2][2], self.v[2][3]))  print('{0:4} {1:4} {2:4} {3:4}'.format(self.v[3][0], self.v[3][1], self.v[3][2], self.v[3][3]))  print('得分为:{0:4}'.format(self.totalScore))  print('游戏是否结束:{0:4}'.format(self.isOver())) #重新排列 def align(self,vList, direction):  for i in range(vList.count(0)):   vList.remove(0)  zeros = [0 for x in range(4-len(vList))]  if direction == 'left':   vList.extend(zeros)  else:   vList[:0] = zeros #将相同的元素相加，返回新增积分 def addSame(self,vList, direction):  increment=0  if direction == 'left':   for i in [0,1,2]:    if vList[i]==vList[i+1] and vList[i+1]!=0:     vList[i] *= 2     vList[i+1] = 0     increment += vList[i]  else:   for i in [3,2,1]:    if vList[i]==vList[i-1] and vList[i-1]!=0:     vList[i] *= 2     vList[i-1] = 0     increment += vList[i]  return increment #处理行和方向,返回新增积分 def handle(self, vList, direction):  self.align(vList, direction)  increment = self.addSame(vList, direction)  self.align(vList, direction)  self.totalScore += increment #直接加到总值  return increment #判断游戏是否结束 def judge(self):     if self.isOver():   print('你输了，游戏结束!')   return False  else:   if self.totalScore >= 2048:    print('你赢了，游戏结束！但是你还可以继续玩。')   return True #判断游戏是否真正结束 def isOver(self):  N = self.calcCharNumber(0)  if N!=0:   return False  else:   for row in range(4):    flag = self.isListOver(self.v[row])    if flag==False:     return False    for col in range(4):    # 将矩阵中一列复制到一个列表中然后处理    vList = [self.v[row][col] for row in range(4)]    flag = self.isListOver(vList)    if flag==False:     return False  return True   #判断一个列表是否还可以合并 def isListOver(self, vList):  for i in [0,1,2]:   if vList[i]==vList[i+1] and vList[i+1]!=0:    return False  return True def calcCharNumber(self, char):  n = 0  for q in self.v:   n += q.count(char)  return n def addElement(self):  # 统计空白区域数目 N  N = self.calcCharNumber(0)  if N!=0:   # 按2和4出现的几率为3/1来产生随机数2和4   num = random.choice([2, 2, 2, 4])    # 产生随机数k，上一步产生的2或4将被填到第k个空白区域   k = random.randrange(1, N+1) #k的范围为[1,N]   n = 0   for i in range(4):    for j in range(4):     if self.v[i][j] == 0:      n += 1      if n == k:       self.v[i][j] = num       return       def moveLeft(self):  self.moveHorizontal('left') def moveRight(self):  self.moveHorizontal('right') def moveHorizontal(self, direction):  for row in range(4):   self.handle(self.v[row], direction)  def moveUp(self):  self.moveVertical('left') def moveDown(self):  self.moveVertical('right') def moveVertical(self, direction):  for col in range(4):   # 将矩阵中一列复制到一个列表中然后处理   vList = [self.v[row][col] for row in range(4)]   self.handle(vList, direction)   # 从处理后的列表中的数字覆盖原来矩阵中的值   for row in range(4):    self.v[row][col] = vList[row]      #主要的处理函数 def operation(self):  op = input('operator:')  if op in ['a', 'A']: # 向左移动   self.moveLeft()   self.addElement()  elif op in ['d', 'D']: # 向右移动   self.moveRight()   self.addElement()  elif op in ['w', 'W']: # 向上移动   self.moveUp()   self.addElement()  elif op in ['s', 'S']: # 向下移动   self.moveDown()   self.addElement()  else:   print('错误的输入。请输入 [W, S, A, D] 或者是其小写')  #开始print('输入：W(上移) S(下移) A(左移) D(右移), press <CR>.')g =game2048()flag = Truewhile True: g.display() flag = g.judge() g.operation() flag = g.judge()

演示图

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