查找所有填充为 0 的矩形的Python程序
我们有一个二维数组,用零和一填充。我们必须找到所有填充为 0 的矩形的起点和终点。假设矩形是分开的并且不相互接触,但是它们可以接触数组的边界。一个矩形可能只包含一个元素。
例子:
input = [
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1]
]
Output:
[
[2, 3, 3, 5], [3, 1, 5, 1], [5, 3, 6, 5]
]
Explanation:
We have three rectangles here, starting from
(2, 3), (3, 1), (5, 3)
Input = [
[1, 0, 1, 1, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 0]
]
Output:
[
[0, 1, 0, 1], [1, 2, 1, 2], [2, 3, 3, 5],
[3, 1, 4, 1], [5, 3, 5, 6], [7, 2, 7, 2],
[7, 6, 7, 6]
]
第 1 步:逐行和逐列查找 0
步骤 2:当遇到任何 0 时,将其位置保存在输出数组中,并使用循环将所有与该位置相关的 0 更改为任意公共数字,以便我们下次将其排除在处理之外。
第 3 步:当您在第 2 步中更改所有相关的 0 时,将最后处理的 0 的位置存储在输出数组中的同一索引中。
第 4 步:当你触摸边缘时要特别小心,不要减去 -1,因为循环在确切的位置断裂。
以下是上述方法的实现:
Python3
# Python program to find all
# rectangles filled with 0
def findend(i,j,a,output,index):
x = len(a)
y = len(a[0])
# flag to check column edge case,
# initializing with 0
flagc = 0
# flag to check row edge case,
# initializing with 0
flagr = 0
for m in range(i,x):
# loop breaks where first 1 encounters
if a[m][j] == 1:
flagr = 1 # set the flag
break
# pass because already processed
if a[m][j] == 5:
pass
for n in range(j, y):
# loop breaks where first 1 encounters
if a[m][n] == 1:
flagc = 1 # set the flag
break
# fill rectangle elements with any
# number so that we can exclude
# next time
a[m][n] = 5
if flagr == 1:
output[index].append( m-1)
else:
# when end point touch the boundary
output[index].append(m)
if flagc == 1:
output[index].append(n-1)
else:
# when end point touch the boundary
output[index].append(n)
def get_rectangle_coordinates(a):
# retrieving the column size of array
size_of_array = len(a)
# output array where we are going
# to store our output
output = []
# It will be used for storing start
# and end location in the same index
index = -1
for i in range(0,size_of_array):
for j in range(0, len(a[0])):
if a[i][j] == 0:
# storing initial position
# of rectangle
output.append([i, j])
# will be used for the
# last position
index = index + 1
findend(i, j, a, output, index)
print (output)
# driver code
tests = [
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1]
]
get_rectangle_coordinates(tests)
输出:
[[2, 3, 3, 5], [3, 1, 5, 1], [5, 3, 6, 5]]
有关详细信息,请参阅有关查找所有填充为 0 的矩形的完整文章!