力扣 85. 最大矩形

动态规划

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
def maximalRectangle(matrix):
if len(matrix) < 1 or len(matrix[0]) < 1:
return 0

m, n = len(matrix), len(matrix[0])

dp = [[0] * n for _ in range(m)]

for i in range(m):
for j in range(n):
if j == 0 and matrix[i][j] == "1":
dp[i][j] = 1
elif matrix[i][j] == "0":
dp[i][j] = 0
elif matrix[i][j] == "1":
dp[i][j] = dp[i][j - 1] + 1
# print(dp)
# 可转化为84. 柱状图中最大的矩形
ret = 0
for i in range(m):
for j in range(n):
if matrix[i][j] == "0":
continue
width, area = dp[i][j], dp[i][j]
for k in range(i - 1, -1, -1):
width = min(width, dp[k][j])
area = max(area, (i - k + 1) * width)
ret = max(ret, area)
return ret

单调栈

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
def maximalRectangle(matrix):
if len(matrix) < 1 or len(matrix[0]) < 1:
return 0

m, n = len(matrix), len(matrix[0])

dp = [[0] * n for _ in range(m)]

for i in range(m):
for j in range(n):
if j == 0 and matrix[i][j] == "1":
dp[i][j] = 1
elif matrix[i][j] == "0":
dp[i][j] = 0
elif matrix[i][j] == "1":
dp[i][j] = dp[i][j - 1] + 1
# print(dp)
# 可转化为84. 柱状图中最大的矩形
ret = 0
for j in range(n):
left, right = [0] * m, [0] * m

tmp = []
for i in range(m):
while tmp and dp[tmp[-1]][j] >= dp[i][j]:
tmp.pop()
left[i] = tmp[-1] if tmp else -1
tmp.append(i)
tmp = []
for i in range(m - 1, -1, -1):
while tmp and dp[tmp[-1]][j] >= dp[i][j]:
tmp.pop()
right[i] = tmp[-1] if tmp else m
tmp.append(i)

for i in range(m):
ret = max(ret, (right[i] - left[i] - 1) * dp[i][j])

return ret
  • Copyright: Copyright is owned by the author. For commercial reprints, please contact the author for authorization. For non-commercial reprints, please indicate the source.
  • Copyrights © 2022 eightyninth
  • Visitors: | Views:

请我喝杯咖啡吧~

支付宝
微信