2048的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 = True while True: g.display() flag = g.judge() g.operation() flag = g.judge()

演示图

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