LeetCode 221. Maximal Square

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题目

Given a 2D binary matrix filled with 0’s and 1’s, find the largest square containing only 1’s and return its area.

Example:

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Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

思路

DP.
两种DP方法（实际是一种）.

  1. dp(i,j) represents the side length of the maximum square whose bottom right corner is the cell with index (i,j) in the original matrix.
    dp(i,j)=min(dp(i−1,j),dp(i−1,j−1),dp(i,j−1))+1.
    Time complexity : O(mn)
    Space complexity : O(mn)

  2. In the previous approach for calculating dp of $i^{th}$row we are using only the previous element and the $(i-1)^{th}$row. Therefore, we don’t need 2D dp matrix as 1D dp array will be sufficient for this.
    dp[j]=min(dp[j-1],dp[j],prev), prev = old dp[j-1].
    Time complexity : O(mn)
    Space complexity : O(n)

代码

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# # 2d DP: dp(i,j)=min(dp(i−1,j),dp(i−1,j−1),dp(i,j−1))+1.

# class Solution(object):
#     def maximalSquare(self, matrix):
#         """
#         :type matrix: List[List[str]]
#         :rtype: int
#         """
#         if matrix == []:
#             return 0
#         res = 0
#         m, n = len(matrix), len(matrix[0])
#         dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
#         for i in range(1, m+1):
#             for j in range(1, n+1):
#                 if matrix[i-1][j-1] == '1':
#                     dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])+1
#                     res = max(res, dp[i][j])
#         return res*res


# # 1d DP: dp[j]=min(dp[j−1],dp[j],prev), prev = dp[j-1]

class Solution(object):
    def maximalSquare(self, matrix):
        """
        :type matrix: List[List[str]]
        :rtype: int
        """
        if matrix == []:
            return 0
        res, prev = 0, 0
        m, n = len(matrix), len(matrix[0])
        dp = [0] * (n+1)
        for i in range(1, m+1):
            for j in range(1, n+1):
                tmp = dp[j]
                if matrix[i-1][j-1] == '1':
                    dp[j] = min(dp[j], dp[j-1], prev)+1
                    res = max(res, dp[j])
                else:
                    dp[j] = 0
                prev = tmp
        return res*res