用递归统计矩阵的连通个数

 

 

def getConnectedNum(mat):
    cnt = 0
    h = len(mat)
    for i in range(h):
        w = len(mat[i])
        for j in range(w):
            if mat[i][j] == 0:
                cnt = cnt + 1
                seekConnected(mat, i, j, h, w)
　　　　　　　　　 # 打印
                for i in range(h):
                    for j in range(w):
                        print(mat[i][j], end="\t")
                    print()
                print()
    return cnt


def seekConnected(mat, i, j, h, w, border = None):
    if i < 0 or j < 0 or i > h-1 or j > w-1 or mat[i][j] == 1:
        return
    mat[i][j] = 1
    seekConnected(mat, i - 1, j, h, w)
    seekConnected(mat, i + 1, j, h, w)
    seekConnected(mat, i, j - 1, h, w)
    seekConnected(mat, i, j + 1, h, w)
    pass

if __name__ == '__main__':
    mat = [
        [0, 0, 1, 1, 0, 1, 1, 1, 1],
        [0, 0, 1, 1, 0, 1, 1, 1, 1],
        [0, 0, 1, 1, 1, 1, 1, 0, 0],
        [0, 0, 1, 1, 0, 1, 1, 1, 1]
    ]

    print(getConnectedNum(mat))
    pass