230821 Surround the 1's
【题意】：给你一个矩阵，矩阵元素值仅有 0 和 1 组成。定义一个元素周围有 8 个元素。求：有多少个值为 1 的元素，它周围值为 0 的元素个数为偶数个。
【Excepted】
- Time Complexity: O(n * m)
- Space Complexity: O(1)
Python3
【玩具❌超时 751 /1120】
py
from functools import reduce
class Solution:
def Count(self, matrix):
n, m= len(matrix), len(matrix[0])
for i in range(n):
matrix[i] = [1, *matrix[i], 1]
matrix = [
[1] * (m+2),
*matrix,
[1] * (m+2),
]
ans = 0
for i in range(1, n+1):
for j in range(1, m+1):
if matrix[i][j] == 0:
continue
r = list(map(lambda x: i+x,
[-1, -1, -1,
0, 0,
1, 1, 1]
))
c = list(map(lambda x: j+x,
[-1, 0, 1,
-1, 1,
-1, 0, 1]
))
def func(count, k):
return count + int(not matrix[r[k]][c[k]])
count = reduce(func, range(8), 0)
ans += 1 if count != 0 and count % 2 == 0 else 0
return ans【改写评论区】
py
class Solution:
def Count(self, matrix):
n, m= len(matrix), len(matrix[0])
ans = 0
for i in range(n):
for j in range(m):
if matrix[i][j] == 0:
continue
r = [-1, -1, -1,
0, 0,
1, 1, 1]
c = [-1, 0, 1,
-1, 1,
-1, 0, 1]
count = 0
for k in range(8):
_i = i + r[k]
_j = j + c[k]
if 0<=_i<n and 0<=_j<m and matrix[_i][_j] == 0:
count += 1
ans += 1 if count != 0 and count % 2 == 0 else 0
return anspy
class Solution:
def Count(self, matrix):
n, m= len(matrix), len(matrix[0])
def f(r, c):
num = 0
num += 0 if r < 1 else int(not matrix[r-1][c]) # top
num += 0 if r == n-1 else int(not matrix[r+1][c]) # bottom
num += 0 if c < 1 else int(not matrix[r][c-1]) # left
num += 0 if c == m-1 else int(not matrix[r][c+1]) # right
num += 0 if r < 1 or c < 1 else int(not matrix[r-1][c-1]) # left-top
num += 0 if r == n-1 or c < 1 else int(not matrix[r+1][c-1]) # left-bottom
num += 0 if r < 1 or c == m-1 else int(not matrix[r-1][c+1]) # right-top
num += 0 if r == n-1 or c == m-1 else int(not matrix[r+1][c+1]) # right-bottom
return 1 if num != 0 and num % 2 == 0 else 0
ans = 0
for i in range(n):
for j in range(m):
if matrix[i][j] == 1:
ans += f(i, j)
return anspy
class Solution:
def Count(self, matrix):
n, m= len(matrix), len(matrix[0])
def f(r, c):
num = 0
num += int(not matrix[r-1][c]) if r > 0 else 0 # top
num += int(not matrix[r+1][c]) if r < n-1 else 0 # bottom
num += int(not matrix[r][c-1]) if c > 0 else 0 # left
num += int(not matrix[r][c+1]) if c < m-1 else 0 # right
num += int(not matrix[r-1][c-1]) if r > 0 and c > 0 else 0 # left-top
num += int(not matrix[r+1][c-1]) if r < n-1 and c > 0 else 0 # left-bottom
num += int(not matrix[r-1][c+1]) if r > 0 and c < m-1 else 0 # right-top
num += int(not matrix[r+1][c+1]) if r < n-1 and c < m-1 else 0 # right-bottom
return 1 if num != 0 and num % 2 == 0 else 0
ans = 0
for i in range(n):
for j in range(m):
if matrix[i][j] == 1:
ans += f(i, j)
return ans