# 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 resultRun Time: 152 ms
Tuesday, August 18, 2015
Leetcode 85. Maximal Rectangle
https://leetcode.com/problems/maximal-rectangle/
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