能不能把河内塔半路解决？我做了大量的研究来寻找能够解决用户配置问题的代码，但是我还没有找到一个。这是一个任务，我需要代码接管从用户已经停止求解的地方，并继续为用户解决它，而不重置为方块一。在

我知道有递归算法在那里随时可用，但这不是我要寻找的。

我正在寻找算法，它可以接管用户解决问题的地方，然后继续从那里解决问题。

有什么想法吗？在

到目前为止，我已经提出了一种算法，它将优化的算法(通过递归实现)存储到一个数组中，然后检查用户的输入是否等于数组中找到的任何一个，然后从那里继续求解。但是，问题在于优化算法数组中找不到用户的配置。在

以下是我目前为止的代码(我排除了堆栈.py代码)：def solveHalfway(n, start, end, middle, count):

gameInstance.stackA = [3,2]

gameInstance.stackB = []

gameInstance.stackC = [1]

loopCounter = 0 # initialise loopCounter as 0

moveCounter = 0 # initialise the move index the user is stuck at

indicator = 0 # to indicate whether the user's config equals the solution's config

while loopCounter < arrayOfStacks.size(): # while loopCounter size has not reached the end of arrayOfStacks

if loopCounter != 0 and loopCounter % 3 == 0: # if 3 stacks have been dequeued

moveCounter += 1

if gameInstance.getUserConfig() == tempStack.data: #check whether user's config is equal to the solution's config

indicator += 1

print "User is stuck at move: ", moveCounter #this will be the current move the user is at

while arrayOfStacks.size() != 0: # while not the end of arrayOfStacks

correctMovesStack.push(arrayOfStacks.dequeue()) # add the moves to correctMovesStack

if correctMovesStack.size() == 3: # if 3 stacks have been dequeued

print "Step:", moveCounter , correctMovesStack.data # display the step number plus the correct move to take

moveCounter+=1 # increase move by 1

while correctMovesStack.size() != 0: # if correct moves stack isn't empty

correctMovesStack.pop() # empty the stack

return

else:

while tempStack.size() != 0: # check if tempStack is empty

tempStack.pop() # empty tempStack so that it can be used for the next loop

tempStack.push(arrayOfStacks.dequeue()) #dequeue from arrayOfStacks for a total of 3 times and push it to tempStack

else:

tempStack.push(arrayOfStacks.dequeue()) #dequeue from arrayOfStacks for a total of 3 times and push it to tempStack

loopCounter +=1 # increase loop counter by 1

if indicator == 0:

moveWith3Towers(noOfDisks, stackA, stackC, stackB, count)

print indicator