# T:O(n^2) S:O(n)
class Solution:
# @param matrix, a list of lists of 1 length string
# @return an integer
def maximalRectangle(self, matrix):
if not matrix:
return 0
result = 0
m = len(matrix)
n = len(matrix[0])
L = [0 for _ in xrange(n)]
H = [0 for _ in xrange(n)]
R = [n for _ in xrange(n)]
for i in xrange(m):
left = 0
for j in xrange(n):
if matrix[i][j] == '1':
L[j] = max(L[j], left)
H[j] += 1
else:
L[j] = 0
H[j] = 0
R[j] = n
left = j + 1
right = n
for j in reversed(xrange(n)):
if matrix[i][j] == '1':
R[j] = min(R[j], right)
result = max(result, H[j] * (R[j] - L[j]))
else:
right = j
return result
Run Time: 152 ms
Tuesday, August 18, 2015
Leetcode 85. Maximal Rectangle
https://leetcode.com/problems/maximal-rectangle/
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