找到具有相同值的相邻小区及其相邻小区等答案

【问题标题】：finding neighbor cells and their neighbor cells and so on with same value找到具有相同值的相邻小区及其相邻小区等
【发布时间】：2019-05-16 10:49:26
【问题描述】：

我有一张这样的桌子

6 5 6 6 3 6
2 4 6 8 1 5
3 6 6 6 4 6
7 4 2 6 1 8

并且用户选择了 0,2 (6) 我必须找到该单元格的所有具有相同值的邻居，以及具有相同值的邻居，依此类推。对于这个表，我需要一个这样的列表

['0,2','0,3','1,2','2,1','2,2','2,3','3,3']

注意：表格大小没有限制。

【问题讨论】：

你的尝试有什么问题？
嗨@EnesEren，您尝试过解决方案吗？如果是这样，请与我们分享。（即使它不工作）。见How to Ask。

标签： python-3.x algorithm matrix

【解决方案1】：

为什么不直接创建与矩阵中当前坐标偏移为 1 的所有组合，而丢弃具有任何负坐标的组合？

【讨论】：

【解决方案2】：

一年后我记起了我的问题并回来回答我自己的问题，因为我现在好一点了。这是我的解决方案。

def findNeighbors(gameMap,selectedRow,selectedColumn):
    finalNeighbors = [] #the list that we will store final state of neighbor list
    newNeighbors = [] #the list that we will store the new neighbors in every loop
    newNeighborsClone = [] #the list that we will traverse inside after clearing the new neighbor list
    newNeighbors.append([selectedRow,selectedColumn]) #we already have a known neighbor which is chosed one
    selectedNumber = gameMap[selectedRow][selectedColumn] 
    print("Selected number = " + str(selectedNumber))
    while newNeighbors: #while newNeighbors list has an element inside
        finalNeighbors = finalNeighbors + newNeighbors #add every new neighbor to main neighbor list
        newNeighborsClone = newNeighbors[::] #clone the new neighbors list
        newNeighbors = [] #clearing the new neighbors list
        for row,column in newNeighborsClone:
            if row == 0: #if we're at peak we only need to look for down
                if gameMap[row+1][column] == selectedNumber and [row+1,column] not in finalNeighbors: #and to prevent infinite loops, we have to check if its already in final neighbors list
                    newNeighbors.append([row+1,column])
            elif row == (len(gameMap)-1): #if we're at the bottom we only need to look for up
                if gameMap[row-1][column] == selectedNumber and [row-1,column] not in finalNeighbors:
                    newNeighbors.append([row-1,column])
            else: #if we're somewhere at middle we have to look up and down
                if gameMap[row+1][column] == selectedNumber and [row+1,column] not in finalNeighbors:
                    newNeighbors.append([row+1,column])
                if gameMap[row-1][column] == selectedNumber and [row-1,column] not in finalNeighbors:
                    newNeighbors.append([row-1,column])
            if column == 0: #if we're at the left side we only need to look for right
                if gameMap[row][column+1] == selectedNumber and [row,column+1] not in finalNeighbors: 
                    newNeighbors.append([row,column+1])
            elif column == (len(gameMap[0])-1): #if we're at the right side we only need to look for left
                if gameMap[row][column-1] == selectedNumber and [row,column-1] not in finalNeighbors:
                    newNeighbors.append([row,column-1])
            else: #if we're somewhere at middle
                if gameMap[row][column+1] == selectedNumber and [row,column+1] not in finalNeighbors:
                    newNeighbors.append([row,column+1])
                if gameMap[row][column-1] == selectedNumber and [row,column-1] not in finalNeighbors:
                    newNeighbors.append([row,column-1])
    finalNeighbors = deleteDuplicates(finalNeighbors) #Even after all of if elses, there can be still duplicated elements. Better removing them
    return finalNeighbors

我无法将整个代码复制到这里，我不知道为什么。但这里是分配来完成这项工作的功能。我可以只使用 try except 块而不是所有这些 if else，但是在该分配中禁止使用 try except。所以我不得不这样做。您可以只检查 try 块中的所有四个方面，而不是那么多 ifs

【讨论】：