一、游戲玩法介紹：
24點游戲是兒時玩的主要益智類游戲之一，玩法為：從一副撲克中抽取4張牌，對4張牌使用加減乘除中的任何方法，使計算結果為24。例如，2,3,4,6，通過( ( ( 4 + 6 ) - 2 ) * 3 ) = 24，最快算出24者剩。
二、設計思路：
由于設計到了表達式，很自然的想到了是否可以使用表達式樹來設計程序。本程序的確使用了表達式樹，也是程序最關鍵的環節。簡要概括為：先列出所有表達式的可能性，然后運用表達式樹計算表達式的值。程序中大量的運用了遞歸，各個遞歸式不是很復雜，大家耐心看看，應該是能看懂的
表達式樹：
表達式樹的所有葉子節點均為操作數(operand)，其他節點為運算符(operator)。由于本例中都是二元運算，所以表達式樹是二叉樹。下圖就是一個表達式樹
具體步驟：
1、遍歷所有表達式的可能情況
遍歷分為兩部分，一部分遍歷出操作數的所有可能，然后是運算符的所有可能。全排列的計算采用了遞歸的思想
#返回一個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、根據傳入的表達式的值，構造表達式樹
由于表達式樹的特點，所有操作數均為葉子節點，操作符為非葉子節點，而一個表達式(例如( ( ( 6 + 4 ) - 2 ) * 3 ) = 24) 只有3個運算符，即一顆表達式樹只有3個非葉子節點。所以樹的形狀只有兩種可能，就直接寫死了
#樹節點
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、計算表達式樹的值
也運用了遞歸
#根據兩個數和一個符號，計算值
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式
還是運用了遞歸
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)
#根據兩個數和一個符號，計算值
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])
以上就是本文的全部內容，希望對大家的學習有所幫助，也希望大家多多支持腳本之家。
更多文章、技術交流、商務合作、聯系博主
微信掃碼或搜索：z360901061
微信掃一掃加我為好友
QQ號聯系： 360901061
您的支持是博主寫作最大的動力，如果您喜歡我的文章，感覺我的文章對您有幫助，請用微信掃描下面二維碼支持博主2元、5元、10元、20元等您想捐的金額吧，狠狠點擊下面給點支持吧，站長非常感激您！手機微信長按不能支付解決辦法：請將微信支付二維碼保存到相冊，切換到微信，然后點擊微信右上角掃一掃功能，選擇支付二維碼完成支付。
【本文對您有幫助就好】元
