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LeetCode-Solutions/Python/maximal-square.py at master · pythonpeixun/LeetCode-Solutions · GitHub
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LeetCode-Solutions
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Python
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maximal-square.py
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LeetCode-Solutions
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Python
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maximal-square.py
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# Time: O(n^2)
# Space: O(n)
class
Solution
(
object
):
# @param {character[][]} matrix
# @return {integer}
def
maximalSquare
(
self
,
matrix
):
if
not
matrix
:
return
0
m
,
n
=
len
(
matrix
),
len
(
matrix
[
0
])
size
=
[[
0
for
j
in
xrange
(
n
)]
for
i
in
xrange
(
2
)]
max_size
=
0
for
j
in
xrange
(
n
):
if
matrix
[
0
][
j
]
==
'1'
:
size
[
0
][
j
]
=
1
max_size
=
max
(
max_size
,
size
[
0
][
j
])
for
i
in
xrange
(
1
,
m
):
if
matrix
[
i
][
0
]
==
'1'
:
size
[
i
%
2
][
0
]
=
1
else
:
size
[
i
%
2
][
0
]
=
0
for
j
in
xrange
(
1
,
n
):
if
matrix
[
i
][
j
]
==
'1'
:
size
[
i
%
2
][
j
]
=
min
(
size
[
i
%
2
][
j
-
1
], \
size
[(
i
-
1
)
%
2
][
j
], \
size
[(
i
-
1
)
%
2
][
j
-
1
])
+
1
max_size
=
max
(
max_size
,
size
[
i
%
2
][
j
])
else
:
size
[
i
%
2
][
j
]
=
0
return
max_size
*
max_size
# Time: O(n^2)
# Space: O(n^2)
# DP.
class
Solution2
(
object
):
# @param {character[][]} matrix
# @return {integer}
def
maximalSquare
(
self
,
matrix
):
if
not
matrix
:
return
0
m
,
n
=
len
(
matrix
),
len
(
matrix
[
0
])
size
=
[[
0
for
j
in
xrange
(
n
)]
for
i
in
xrange
(
m
)]
max_size
=
0
for
j
in
xrange
(
n
):
if
matrix
[
0
][
j
]
==
'1'
:
size
[
0
][
j
]
=
1
max_size
=
max
(
max_size
,
size
[
0
][
j
])
for
i
in
xrange
(
1
,
m
):
if
matrix
[
i
][
0
]
==
'1'
:
size
[
i
][
0
]
=
1
else
:
size
[
i
][
0
]
=
0
for
j
in
xrange
(
1
,
n
):
if
matrix
[
i
][
j
]
==
'1'
:
size
[
i
][
j
]
=
min
(
size
[
i
][
j
-
1
], \
size
[
i
-
1
][
j
], \
size
[
i
-
1
][
j
-
1
])
+
1
max_size
=
max
(
max_size
,
size
[
i
][
j
])
else
:
size
[
i
][
j
]
=
0
return
max_size
*
max_size
# Time: O(n^2)
# Space: O(n^2)
# DP.
class
Solution3
(
object
):
# @param {character[][]} matrix
# @return {integer}
def
maximalSquare
(
self
,
matrix
):
if
not
matrix
:
return
0
H
,
W
=
0
,
1
# DP table stores (h, w) for each (i, j).
table
=
[[[
0
,
0
]
for
j
in
xrange
(
len
(
matrix
[
0
]))] \
for
i
in
xrange
(
len
(
matrix
))]
for
i
in
reversed
(
xrange
(
len
(
matrix
))):
for
j
in
reversed
(
xrange
(
len
(
matrix
[
i
]))):
# Find the largest h such that (i, j) to (i + h - 1, j) are feasible.
# Find the largest w such that (i, j) to (i, j + w - 1) are feasible.
if
matrix
[
i
][
j
]
==
'1'
:
h
,
w
=
1
,
1
if
i
+
1
<
len
(
matrix
):
h
=
table
[
i
+
1
][
j
][
H
]
+
1
if
j
+
1
<
len
(
matrix
[
i
]):
w
=
table
[
i
][
j
+
1
][
W
]
+
1
table
[
i
][
j
]
=
[
h
,
w
]
# A table stores the length of largest square for each (i, j).
s
=
[[
0
for
j
in
xrange
(
len
(
matrix
[
0
]))] \
for
i
in
xrange
(
len
(
matrix
))]
max_square_area
=
0
for
i
in
reversed
(
xrange
(
len
(
matrix
))):
for
j
in
reversed
(
xrange
(
len
(
matrix
[
i
]))):
side
=
min
(
table
[
i
][
j
][
H
],
table
[
i
][
j
][
W
])
if
matrix
[
i
][
j
]
==
'1'
:
# Get the length of largest square with bottom-left corner (i, j).
if
i
+
1
<
len
(
matrix
)
and
j
+
1
<
len
(
matrix
[
i
+
1
]):
side
=
min
(
s
[
i
+
1
][
j
+
1
]
+
1
,
side
)
s
[
i
][
j
]
=
side
max_square_area
=
max
(
max_square_area
,
side
*
side
)
return
max_square_area
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