递归到迭代DFS的python语言

Question

我正试图将递归代码转换为迭代代码，任务是找到网格中最大的区域（由1组成的连接单元）。任务是找到网格中最大的区域（由1组成的连接单元）。

代码参考了这里的内容。https:/www.geeksforgeeks.orgfind-length-largest-region-boolean-matrix我试过用堆栈和循环来代替递归，但没有效果。

这是我试过的代码，但它不能重现与递归方法相同的结果。

我已经用

M = np.array([[0, 0, 1, 1, 0], [1, 0, 1, 1, 0], [0, 1, 0, 0, 0], [0, 0, 0, 0, 2],[0, 0, 0, 1, 0],[0, 0, 3, 0, 0]]) 

def DFS(M, row, col, visited): 
    rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1]  
    colNbr = [-1, 0, 1, -1, 1, -1, 0, 1]  

    # Mark this cell as visited  
    visited[row][col] = True
    stack = [] 

    for k in range(8): 
        if (isSafe(M, row + rowNbr[k],  
                   col + colNbr[k], visited)): 

            stack.push(M[row][col])
            row = row + rowNbr[k]
            col = col + colNbr[k]
            for k in range(8): 
                if (isSafe(M, row + rowNbr[k],  
                    col + colNbr[k], visited)):
                    stack.push(M[row][col])

Answer 1

使用堆栈的DFS的一般结构为

stack = [] # initialize Stack
visited = set() # initialize hash table for looking at visited nodes
stack.append(startNode) # put in the start node

while len(stack) != 0: # check whether there is anything in the To-Do list
   newNode = stack.pop() # get next node to visit
   if newNode not in visited: # update visited if this node has not been visited
      visited.add(newNode) 
   for neighbor in newNode.neighbors: # iterate over neighbors
      if neighbor not in visited: # check whether neighbors were visited
         stack.append(neighbor) # this node was not seen before, add it to To-Do list

在你的例子中，似乎你的迭代次数并不取决于堆栈中是否还有一个元素，而且由于你没有从堆栈中取出任何元素，所以你没有从堆栈中得到下一个要访问的元素。

你可以试试下面的方法。

def DFS(M, row, col, visited):
    rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1]
    colNbr = [-1, 0, 1, -1, 1, -1, 0, 1]
    # initialize stack
    stack = [] 
    stack.append((row, col))
    while len(stack) != 0:
        row, col = stack.pop()
        if not visited[row, col]:
            visited[row, col] = 1
        # iterate over neighbors
        for k in range(8):
            if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited)):
                row = row + rowNbr[k]
                col = col + colNbr[k]
                if not visited[row, col]:
                    stack.append((row, col))

这使用了一个元组 (row, col) 来标记位置并将其存储在堆栈上。