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仅包含 0 和 1 的二维矩阵 - python

[英]2D matrix containing only 0s and 1s - python

With a given 2D matrix containing only 0s and 1s.给定的二维矩阵只包含 0 和 1。 Each 0 represents land, and each 1 represents part of a river.每个 0 代表土地,每个 1 代表河流的一部分。 A river consists of any number of 1s that are either horizontally or vertically adjacent (but not diagonally adjacent).一条河流由任意数量的 1 组成,这些 1 水平或垂直相邻(但不是对角相邻)。 The number of adjacent 1s forming a river determine its size.形成一条河流的相邻 1 的数量决定了它的大小。 Write a function that returns an array of the sizes of all rivers represented in the input matrix.编写一个 function,它返回一个数组,该数组包含输入矩阵中表示的所有河流的大小。

   Input:
array =[
        [0, 0, 1, 0, 1, 0],
        [1, 0, 0, 1, 1, 0],
        [1, 0, 0, 0, 0, 1],
        [1, 0, 0, 1, 1, 1],
    ]
output:
There are 4 rivers in the given matrix, and their sizes are: [1, 3, 3, 4]

but actually im expecting to have: [1, 2, 2, 2, 3, 3]但实际上我希望拥有: [1, 2, 2, 2, 3, 3]

because im looking for only rivers which are horizontally or vertically adjacent but not a river which is both.因为我只寻找水平或垂直相邻的河流,而不是两者兼而有之的河流。 (as my algorith output brings). (正如我的算法 output 带来的)。

MY algorithm uses DFS:我的算法使用 DFS:

def riverSizes(matrix):
    sizes = []  # # It would contain all of our river sizes
    visited = [[False for value in row] for row in matrix]
    for i in range(len(matrix)):
        for j in range(len(matrix[i])):
            if visited[i][j]:
                continue
            traverseNode(i, j, matrix, visited, sizes)
    return sizes


def traverseNode(i, j, matrix, visited, sizes):
    currentRiverSize = 0
    nodesToExplore = [[i, j]]
    while len(nodesToExplore):
        currentNode = nodesToExplore.pop()
        i = currentNode[0]
        j = currentNode[1]
        if visited[i][j]:
            continue
        visited[i][j] = True
        if matrix[i][j] == 0:
            continue
        currentRiverSize += 1
        unvisitedNeighbors = getUnvisitedNeighbors(i, j, matrix, visited)
        for neighbor in unvisitedNeighbors:
            nodesToExplore.append(neighbor)
    if currentRiverSize > 0:
        sizes.append(currentRiverSize)


def getUnvisitedNeighbors(i, j, matrix, visited):
    unvisitedNeighbors = []
    if i > 0 and not visited[i - 1][j]:
        unvisitedNeighbors.append([i - 1, j])
    if i < len(matrix) - 1 and not visited[i + 1][j]:
        unvisitedNeighbors.append([i + 1, j])
    if j > 0 and not visited[i][j - 1]:
        unvisitedNeighbors.append([i, j - 1])
    if j < len(matrix[0]) - 1 and not visited[i][j + 1]:
        unvisitedNeighbors.append([i, j + 1])
    return unvisitedNeighbors

how can i fix it?我该如何解决?

The algorithm is giving you the right result for the problem [1, 3, 3, 4].该算法为您提供了问题 [1, 3, 3, 4] 的正确结果。 A river is composed of 1s that are horizontally or vertically adjacent to each other (a 1 can not be horizontally and vertically adjacent to another 1 at the same time, in the case the confusion is the 'either' 'or' of the problem statement)).一条河流由水平或垂直相邻的 1 组成(一个 1 不能同时与另一个 1 水平和垂直相邻,在这种情况下混淆是问题陈述的 'either' 'or' ))。 That is why getUnvisitedNeighbors looks into all vertical and horizontal directions from a field containing a 1. In other words, a river could have an L shape resulting matrix这就是为什么 getUnvisitedNeighbors 从包含 1 的字段中查看所有垂直和水平方向的原因。换句话说,河流可能有一个 L 形状的结果矩阵

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