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 .')g =game2048()flag = truewhile true: g.display() flag = g.judge() g.operation() flag = g.judge()
演示图
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