一、游戏玩法介绍：
24点游戏是儿时玩的主要益智类游戏之一，玩法为：从一副扑克中抽取4张牌，对4张牌使用加减乘除中的任何方法，使计算结果为24。例如，2,3,4,6，通过( ( ( 4 + 6 ) - 2 ) * 3 ) = 24，最快算出24者剩。
二、设计思路：
由于设计到了表达式，很自然的想到了是否可以使用表达式树来设计程序。本程序的确使用了表达式树，也是程序最关键的环节。简要概括为：先列出所有表达式的可能性，然后运用表达式树计算表达式的值。程序中大量的运用了递归，各个递归式不是很复杂，大家耐心看看，应该是能看懂的
表达式树：
表达式树的所有叶子节点均为操作数(operand)，其他节点为运算符(operator)。由于本例中都是二元运算，所以表达式树是二叉树。下图就是一个表达式树
具体步骤：
1、遍历所有表达式的可能情况
遍历分为两部分，一部分遍历出操作数的所有可能，然后是运算符的所有可能。全排列的计算采用了递归的思想
#返回一个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、根据传入的表达式的值，构造表达式树
由于表达式树的特点，所有操作数均为叶子节点，操作符为非叶子节点，而一个表达式(例如( ( ( 6 + 4 ) - 2 ) * 3 ) = 24) 只有3个运算符，即一颗表达式树只有3个非叶子节点。所以树的形状只有两种可能，就直接写死了
#树节点
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、计算表达式树的值
也运用了递归
#根据两个数和一个符号，计算值
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、输出所有可能的表达式
还是运用了递归
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、输出结果
三、所有源码
#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)
#根据两个数和一个符号，计算值
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
#返回一个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])
时间： 2024-10-08 23:02:14