二维矩阵求最大连通区域面积(python)

⼆维矩阵求最⼤连通区域⾯积（python）

2020校招远景的编程题，⼆维01矩阵，横竖算相连，最⼤的连通区域。

思路：⽤⼀个visited列表保存已经过的点，对没的点，如果是1，⽤递归⽅式按上下左右四个⽅向可以连在⼀起的点，存⼊⼀个列表，不断⽐较这个列表长度和之前保存的最⼤值来更新最⼤区域⾯积。

import sys

M,N = list(map(int,adline().strip().split(",")))

mat = []

for _ in range(M):

line = list(map(int,adline().strip().split(",")))

mat.append(line)

def go(i,j,f):

black = []

if 1 in f:

if j>0 and mat[i][j-1] == 1 and [i,j-1] not in visited:

black.append([i,j-1])

visited.append([i,j-1])

利路防水接头

倒立摆if 2 in f:

if i>0 and mat[i-1][j] == 1 and [i-1,j] not in visited:

black.append([i-1,j])

visited.append([i-1,j])

if 3 in f:

多媒体调度台if j<N-1 and mat[i][j+1] == 1 and [i,j+1] not in visited:连通区域

black.append([i,j+1])

visited.append([i,j+1])

网络滤波器

if 4 in f:

if i<M-1 and mat[i+1][j] == 1 and [i+1,j] not in visited:

black.append([i+1,j])

visited.append([i+1,j])

return black

visited = []

res = 0

for i in range(M):

for j in range(N):

if mat[i][j] == 1 and [i,j] not in visited:

visited.append([i,j])

s = go(i,j,[3,4])

if len(s)>res:

res = len(s)+1

print(res)

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标签：保存列表防水

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