- 汉诺塔问题的非递归解法（python语言类解法）
#!/usr/bin/env python
#coding:utf-8
import sys
import time
reload(sys)
sys.setdefaultencoding('utf-8')
class Mycolumns(object):
val=0
#__slots__ = ['plates','name']
def __init__(self,name='',plates_num=0):
self.name = name
self.plates = []
if plates_num > 0 :
for i in range(0,plates_num):
self.plates.append(n-i)
@staticmethod
def fun():
Mycolumns.val +=1
print"this is the %d th time to move" %(Mycolumns.val)
【这段可以用类方法代替】
【
@classmethod
def fun(cls):
cls.val +=1
print"this is the %d th time to move" %(cls.val)
】
def initialize(n):
stack = []
mycolumn1 = Mycolumns('A',n)
if n%2 == 0:
mycolumn2 = Mycolumns('B')
mycolumn3 = Mycolumns('C')
index = 2
else:
mycolumn2 = Mycolumns('C')
mycolumn3 = Mycolumns('B')
index = 1
stack.append(mycolumn1)
stack.append(mycolumn2)
stack.append(mycolumn3)
return stack,index
def nowcolumn(i,stack):
for item in stack:
if i in item.plates:
return item
def nextcolumn(i,stack):
for item in stack:
if i in item.plates:
if i%2!=0:
next = (stack.index(item)+1)%3
else:
next = (stack.index(item)+2)%3
#print "%d next column is %s"%(i,stack[next].name)
return stack[next]
def move(nowcolumn,nextcolumn):
n = nowcolumn.plates.pop()
nextcolumn.plates.append(n)
print "move plate %d from %s to %s" %(n,nowcolumn.name,nextcolumn.name)
Mycolumns.fun()
def hannuoyi(n):
stack,index = initialize(n)
#max = pow(2,n)-1
#k =0
#while(k<max):
FINAL = []
for i in range(0,n):
FINAL.append(n-i)
while(stack[index].plates!=FINAL):
for i in range(1,n+1):
#print "i value is %d" %(i)
if(nowcolumn(i,stack).plates.index(i)==len(nowcolumn(i,stack).plates)-1 and (nextcolumn(i,stack).plates==[] or i< nextcolumn(i,stack).plates[-1])):
move(nowcolumn(i,stack),nextcolumn(i,stack))
#k = k+1
else:
pass
#print"can not move plate %d" %(i)
print stack[0].plates,stack[0].name
print stack[1].plates,stack[1].name
print stack[2].plates,stack[2].name
if __name__=="__main__":
n=3
hannuoyi(n)
2. 汉诺塔问题的非递归解法（python语言过程式解法）
#!/usr/bin/env python
#coding:utf-8
import sys
import time
reload(sys)
sys.setdefaultencoding('utf-8')
global a
a =0
def fun():
global a
a = a+1
print"this is the %d th time to move" %(a)
def nowcolumn(i,stackA,stackB,stackC):
if i in stackA:
return stackA
elif i in stackB:
return stackB
else:
return stackC
def nextcolumn(n,i,stackA,stackB,stackC):
if n%2==0:
if i in stackA:
if i%2!=0:
newcolumn = stackB
else:
newcolumn = stackC
if i in stackB:
if i%2!=0:
newcolumn = stackC
else:
newcolumn = stackA
if i in stackC:
if i%2!=0:
newcolumn = stackA
else:
newcolumn = stackB
else:
if i in stackA:
if i%2==0:
newcolumn = stackB
else:
newcolumn = stackC
if i in stackB:
if i%2==0:
newcolumn = stackC
else:
newcolumn = stackA
if i in stackC:
if i%2==0:
newcolumn = stackA
else:
newcolumn = stackB
return newcolumn
def move(nowcolumn,nextcolumn):
n = nowcolumn.pop()
nextcolumn.append(n)
print "move plate %d" %(n)
fun()
def hannuoyi(n):
stackA = []
stackB = []
stackC = []
FINAL = []
for i in range(0,n):
stackA.append(n-i)
FINAL.append(n-i)
print stackA,stackB,stackC,FINAL
while(stackC!=FINAL):
for i in range(1,n+1):
print "i value is %d" %(i)
if(nowcolumn(i,stackA,stackB,stackC).index(i)==len(nowcolumn(i,stackA,stackB,stackC))-1 and (nextcolumn(n,i,stackA,stackB,stackC)==[] or i< nextcolumn(n,i,stackA,stackB,stackC)[-1])):
move(nowcolumn(i,stackA,stackB,stackC),nextcolumn(n,i,stackA,stackB,stackC))
else:
print"can not move plate %d" %(i)
print stackA,stackB,stackC
if __name__=="__main__":
n=6
hannuoyi(n)
- 汉诺塔问题的递归解法（python语言）
#!/usr/bin/env python
#coding:utf-8
import sys
reload(sys)
sys.setdefaultencoding('utf-8')
global a
a =0
def fun():
global a
a = a+1
def hannuoyi(n,A,B,C):
if n==1:
move(1,A,C)
else:
hannuoyi(n-1,A,C,B)
move(n,A,C)
hannuoyi(n-1,B,A,C)
def move(n,tempA,tempB):
print "move plate %d from column %s to column %s" %(n,tempA,tempB)
fun()
if __name__=="__main__":
hannuoyi(3,'A','B','C')
print "total:we need %d steps"%(a)