python如何實(shí)現(xiàn)24點(diǎn)游戲程序-創(chuàng)新互聯(lián)
這篇文章主要介紹python如何實(shí)現(xiàn)24點(diǎn)游戲程序，文中介紹的非常詳細(xì)，具有一定的參考價(jià)值，感興趣的小伙伴們一定要看完！
一、游戲玩法介紹：
24點(diǎn)游戲是兒時(shí)玩的主要益智類(lèi)游戲之一，玩法為：從一副撲克中抽取4張牌，對(duì)4張牌使用加減乘除中的任何方法，使計(jì)算結(jié)果為24。例如，2,3,4,6，通過(guò)( ( ( 4 + 6 ) - 2 ) * 3 ) = 24，最快算出24者剩。
二、設(shè)計(jì)思路：
由于設(shè)計(jì)到了表達(dá)式，很自然的想到了是否可以使用表達(dá)式樹(shù)來(lái)設(shè)計(jì)程序。本程序的確使用了表達(dá)式樹(shù)，也是程序最關(guān)鍵的環(huán)節(jié)。簡(jiǎn)要概括為：先列出所有表達(dá)式的可能性，然后運(yùn)用表達(dá)式樹(shù)計(jì)算表達(dá)式的值。程序中大量的運(yùn)用了遞歸，各個(gè)遞歸式不是很復(fù)雜，大家耐心看看，應(yīng)該是能看懂的
表達(dá)式樹(shù)：
表達(dá)式樹(shù)的所有葉子節(jié)點(diǎn)均為操作數(shù)(operand)，其他節(jié)點(diǎn)為運(yùn)算符(operator)。由于本例中都是二元運(yùn)算，所以表達(dá)式樹(shù)是二叉樹(shù)。下圖就是一個(gè)表達(dá)式樹(shù)
具體步驟：
1、遍歷所有表達(dá)式的可能情況
遍歷分為兩部分，一部分遍歷出操作數(shù)的所有可能，然后是運(yùn)算符的所有可能。全排列的計(jì)算采用了遞歸的思想
#返回一個(gè)列表的全排列的列表集合 def list_result(l): if len(l) == 1: return [l] all_result = [] for index,item in enumerate(l): r = list_result(l[0:index] + l[index+1:]) map(lambda x : x.append(item),r) all_result.extend(r) return all_result
2、根據(jù)傳入的表達(dá)式的值，構(gòu)造表達(dá)式樹(shù)
由于表達(dá)式樹(shù)的特點(diǎn)，所有操作數(shù)均為葉子節(jié)點(diǎn)，操作符為非葉子節(jié)點(diǎn)，而一個(gè)表達(dá)式(例如( ( ( 6 + 4 ) - 2 ) * 3 ) = 24) 只有3個(gè)運(yùn)算符，即一顆表達(dá)式樹(shù)只有3個(gè)非葉子節(jié)點(diǎn)。所以樹(shù)的形狀只有兩種可能，就直接寫(xiě)死了
#樹(shù)節(jié)點(diǎn) class Node: def __init__(self, val): self.val = val self.left = None self.right = None
def one_expression_tree(operators, operands): root_node = Node(operators[0]) operator1 = Node(operators[1]) operator2 = Node(operators[2]) operand0 = Node(operands[0]) operand1 = Node(operands[1]) operand2 = Node(operands[2]) operand3 = Node(operands[3]) root_node.left = operator1 root_node.right =operand0 operator1.left = operator2 operator1.right = operand1 operator2.left = operand2 operator2.right = operand3 return root_node def two_expression_tree(operators, operands): root_node = Node(operators[0]) operator1 = Node(operators[1]) operator2 = Node(operators[2]) operand0 = Node(operands[0]) operand1 = Node(operands[1]) operand2 = Node(operands[2]) operand3 = Node(operands[3]) root_node.left = operator1 root_node.right =operator2 operator1.left = operand0 operator1.right = operand1 operator2.left = operand2 operator2.right = operand3 return root_node
3、計(jì)算表達(dá)式樹(shù)的值
也運(yùn)用了遞歸
#根據(jù)兩個(gè)數(shù)和一個(gè)符號(hào)，計(jì)算值 def cal(a, b, operator): return operator == '+' and float(a) + float(b) or operator == '-' and float(a) - float(b) or operator == '*' and float(a) * float(b) or operator == '÷' and float(a)/float(b) def cal_tree(node): if node.left is None: return node.val return cal(cal_tree(node.left), cal_tree(node.right), node.val)
4、輸出所有可能的表達(dá)式
還是運(yùn)用了遞歸
def print_expression_tree(root):
print_node(root)
print ' = 24'
def print_node(node):
if node is None :
return
if node.left is None and node.right is None:
print node.val,
else:
print '(',
print_node(node.left)
print node.val,
print_node(node.right)
print ')',
#print ' ( %s %s %s ) ' % (print_node(node.left), node.val, print_node(node.right)),5、輸出結(jié)果
三、所有源碼
#coding:utf-8
from __future__ import division
from Node import Node
def calculate(nums):
nums_possible = list_result(nums)
operators_possible = list_result(['+','-','*','÷'])
goods_noods = []
for nums in nums_possible:
for op in operators_possible:
node = one_expression_tree(op, nums)
if cal_tree(node) == 24:
goods_noods.append(node)
node = two_expression_tree(op, nums)
if cal_tree(node) == 24:
goods_noods.append(node)
map(lambda node: print_expression_tree(node), goods_noods)
def cal_tree(node):
if node.left is None:
return node.val
return cal(cal_tree(node.left), cal_tree(node.right), node.val)
#根據(jù)兩個(gè)數(shù)和一個(gè)符號(hào)，計(jì)算值
def cal(a, b, operator):
return operator == '+' and float(a) + float(b) or operator == '-' and float(a) - float(b) or operator == '*' and float(a) * float(b) or operator == '÷' and float(a)/float(b)
def one_expression_tree(operators, operands):
root_node = Node(operators[0])
operator1 = Node(operators[1])
operator2 = Node(operators[2])
operand0 = Node(operands[0])
operand1 = Node(operands[1])
operand2 = Node(operands[2])
operand3 = Node(operands[3])
root_node.left = operator1
root_node.right =operand0
operator1.left = operator2
operator1.right = operand1
operator2.left = operand2
operator2.right = operand3
return root_node
def two_expression_tree(operators, operands):
root_node = Node(operators[0])
operator1 = Node(operators[1])
operator2 = Node(operators[2])
operand0 = Node(operands[0])
operand1 = Node(operands[1])
operand2 = Node(operands[2])
operand3 = Node(operands[3])
root_node.left = operator1
root_node.right =operator2
operator1.left = operand0
operator1.right = operand1
operator2.left = operand2
operator2.right = operand3
return root_node
#返回一個(gè)列表的全排列的列表集合
def list_result(l):
if len(l) == 1:
return [l]
all_result = []
for index,item in enumerate(l):
r = list_result(l[0:index] + l[index+1:])
map(lambda x : x.append(item),r)
all_result.extend(r)
return all_result
def print_expression_tree(root):
print_node(root)
print ' = 24'
def print_node(node):
if node is None :
return
if node.left is None and node.right is None:
print node.val,
else:
print '(',
print_node(node.left)
print node.val,
print_node(node.right)
print ')',
if __name__ == '__main__':
calculate([2,3,4,6])以上是“python如何實(shí)現(xiàn)24點(diǎn)游戲程序”這篇文章的所有內(nèi)容，感謝各位的閱讀！希望分享的內(nèi)容對(duì)大家有幫助，更多相關(guān)知識(shí)，歡迎關(guān)注創(chuàng)新互聯(lián)行業(yè)資訊頻道！
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