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这是一个Python3的实现的2048,感觉挺有意思的,上截图
程序使用Python3 写的,代码150行左右,基于控制台,方向键使用输入字符模拟。
- # -*- coding:UTF-8 -*-
- #! /usr/bin/python3
- import random
- v = [[0, 0, 0, 0],
- [0, 0, 0, 0],
- [0, 0, 0, 0],
- [0, 0, 0, 0]]
- def display(v, score):
- '''显示界面
- '''
- print('{0:4} {1:4} {2:4} {3:4}'.format(v[0][0], v[0][1], v[0][2], v[0][3]))
- print('{0:4} {1:4} {2:4} {3:4}'.format(v[1][0], v[1][1], v[1][2], v[1][3]))
- print('{0:4} {1:4} {2:4} {3:4}'.format(v[2][0], v[2][1], v[2][2], v[2][3]))
- print('{0:4} {1:4} {2:4} {3:4}'.format(v[3][0], v[3][1], v[3][2], v[3][3]), ' Total score: ', score)
- def init(v):
- '''随机分布网格值
-
- '''
- for i in range(4):
- v[i] = [random.choice([0, 0, 0, 2, 2, 4]) for x in range(4)]
- def align(vList, direction):
- '''对齐非零的数字
- direction == 'left':向左对齐,例如[8,0,0,2]左对齐后[8,2,0,0]
- direction == 'right':向右对齐,例如[8,0,0,2]右对齐后[0,0,8,2]
- '''
- # 移除列表中的0
- for i in range(vList.count(0)):
- vList.remove(0)
- # 被移除的0
- zeros = [0 for x in range(4 - len(vList))]
- # 在非0数字的一侧补充0
- if direction == 'left':
- vList.extend(zeros)
- else:
- vList[:0] = zeros
-
- def addSame(vList, direction):
- '''在列表查找相同且相邻的数字相加, 找到符合条件的返回True,否则返回False,同时还返回增加的分数
-
- direction == 'left':从右向左查找,找到相同且相邻的两个数字,左侧数字翻倍,右侧数字置0
- direction == 'right':从左向右查找,找到相同且相邻的两个数字,右侧数字翻倍,左侧数字置0
- '''
- score = 0
- if direction == 'left':
- for i in [0, 1, 2]:
- if vList[i] == vList[i+1] != 0:
- vList[i] *= 2
- vList[i+1] = 0
- score += vList[i]
- return {'bool':True, 'score':score}
- else:
- for i in [3, 2, 1]:
- if vList[i] == vList[i-1] != 0:
- vList[i-1] *= 2
- vList[i] = 0
- score += vList[i-1]
- return {'bool':True, 'score':score}
- return {'bool':False, 'score':score}
- def handle(vList, direction):
- '''处理一行(列)中的数据,得到最终的该行(列)的数字状态值, 返回得分
- vList: 列表结构,存储了一行(列)中的数据
- direction: 移动方向,向上和向左都使用方向'left',向右和向下都使用'right'
- '''
- totalScore = 0
- align(vList, direction)
- result = addSame(vList, direction)
- while result['bool'] == True:
- totalScore += result['score']
- align(vList, direction)
- result = addSame(vList, direction)
- return totalScore
-
- def operation(v):
- '''根据移动方向重新计算矩阵状态值,并记录得分
- '''
- totalScore = 0
- gameOver = False
- direction = 'left'
- op = input('operator:')
- if op in ['a', 'A']: # 向左移动
- direction = 'left'
- for row in range(4):
- totalScore += handle(v[row], direction)
- elif op in ['d', 'D']: # 向右移动
- direction = 'right'
- for row in range(4):
- totalScore += handle(v[row], direction)
- elif op in ['w', 'W']: # 向上移动
- direction = 'left'
- for col in range(4):
- # 将矩阵中一列复制到一个列表中然后处理
- vList = [v[row][col] for row in range(4)]
- totalScore += 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):
- v[row][col] = vList[row]
- elif op in ['s', 'S']: # 向下移动
- direction = 'right'
- for col in range(4):
- # 同上
- vList = [v[row][col] for row in range(4)]
- totalScore += handle(vList, direction)
- for row in range(4):
- v[row][col] = vList[row]
- else:
- print('Invalid input, please enter a charactor in [W, S, A, D] or the lower')
- return {'gameOver':gameOver, 'score':totalScore}
- # 统计空白区域数目 N
- N = 0
- for q in v:
- N += q.count(0)
- # 不存在剩余的空白区域时,游戏结束
- if N == 0:
- gameOver = True
- return {'gameOver':gameOver, 'score':totalScore}
- # 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)
- n = 0
- for i in range(4):
- for j in range(4):
- if v[i][j] == 0:
- n += 1
- if n == k:
- v[i][j] = num
- break
- return {'gameOver':gameOver, 'score':totalScore}
- init(v)
- score = 0
- print('Input:W(Up) S(Down) A(Left) D(Right), press <CR>.')
- while True:
- display(v, score)
- result = operation(v)
- if result['gameOver'] == True:
- print('Game Over, You failed!')
- print('Your total score:', score)
- else:
- score += result['score']
- if score >= 2048:
- print('Game Over, You Win!!!')
- print('Your total score:', score)
复制代码
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