能不能把河內塔半路解決？我做了大量的研究來尋找能夠解決用戶配置問題的代碼，但是我還沒有找到一個。這是一個任務，我需要代碼接管從用戶已經停止求解的地方，并繼續為用戶解決它，而不重置為方塊一。在

我知道有遞歸算法在那里隨時可用，但這不是我要尋找的。

我正在尋找算法，它可以接管用戶解決問題的地方，然后繼續從那里解決問題。

有什么想法嗎？在

到目前為止，我已經提出了一種算法，它將優化的算法(通過遞歸實現)存儲到一個數組中，然后檢查用戶的輸入是否等于數組中找到的任何一個，然后從那里繼續求解。但是，問題在于優化算法數組中找不到用戶的配置。在

以下是我目前為止的代碼(我排除了堆棧.py代碼)：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

總結

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