一個2048小遊戲的python實現 python
今天看了OSC網友xiaohui_hubei的2048遊戲代碼感受頗有意思,特地花時間玩了一下。發現有兩個小bug吧。 dom
主要有兩個: 函數
1. 原版遊戲每次只消除一次,而不是遞歸消除。如 [2 ,2 ,2 ,2] 左移動的話應該是 [4, 4, 0, 0] , 而不是[8 , 0 , 0 ,0]
2. 對遊戲結束的偵測有bug,沒有那麼簡單。 ui
我從新用OOP的思想重寫了一遍,改正了bug。 spa
# -*- 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('得分爲:{0:4}'.format(self.totalScore)) print('遊戲是否結束:{0:4}'.format(self.isOver())) #從新排列 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 #將相同的元素相加,返回新增積分 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 #處理行和方向,返回新增積分 def handle(self, vList, direction): self.align(vList, direction) increment = self.addSame(vList, direction) self.align(vList, direction) self.totalScore += increment #直接加到總值 return increment #判斷遊戲是否結束 def judge(self): if self.isOver(): print('你輸了,遊戲結束!') return False else: if self.totalScore >= 2048: print('你贏了,遊戲結束!可是你還能夠繼續玩。') return True #判斷遊戲是否真正結束 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): # 將矩陣中一列複製到一個列表中而後處理 vList = [self.v[row][col] for row in range(4)] flag = self.isListOver(vList) if flag==False: return False return True #判斷一個列表是否還能夠合併 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): # 統計空白區域數目 N N = self.calcCharNumber(0) if N!=0: # 按2和4出現的概率爲3/1來產生隨機數2和4 num = random.choice([2, 2, 2, 4]) # 產生隨機數k,上一步產生的2或4將被填到第k個空白區域 k = random.randrange(1, N+1) #k的範圍爲[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): # 將矩陣中一列複製到一個列表中而後處理 vList = [self.v[row][col] for row in range(4)] self.handle(vList, direction) # 從處理後的列表中的數字覆蓋原來矩陣中的值 for row in range(4): self.v[row][col] = vList[row] #主要的處理函數 def operation(self): op = input('operator:') if op in ['a', 'A']: # 向左移動 self.moveLeft() self.addElement() elif op in ['d', 'D']: # 向右移動 self.moveRight() self.addElement() elif op in ['w', 'W']: # 向上移動 self.moveUp() self.addElement() elif op in ['s', 'S']: # 向下移動 self.moveDown() self.addElement() else: print('錯誤的輸入。請輸入 [W, S, A, D] 或者是其小寫') #開始 print('輸入: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()