leetcode 221. Maximal Square （python）

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描述

Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example 1:

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 4
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Example 2:

Input: matrix = [["0","1"],["1","0"]]
Output: 1
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Note:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j] is '0' or '1'.
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解析

根据题意，给出了一个 m x n 的矩阵 matrix ，里面每个格子填充了 0 或者 1 ，找到只包含 1 的最大的正方形面积并返回。其实这道题一看完题目，我们就能知道这种数数的题目基本上用动态规划会比较方便一点，我们定义 dp[i][j] 为以此位置做右下角的正方形的最大边长，公式为：

dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
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用图展示会更加清楚：

最后只需要找到 dp 中的最大值作为正方形边长，得到正方形的最大面积即可。其实如果是经常刷题的朋友应该知道，这道题和 1277. Count Square Submatrices with All Ones 几乎一样。

解答

import numpy
class Solution(object):
    def maximalSquare(self, matrix):
        """
        :type matrix: List[List[str]]
        :rtype: int
        """
        M = len(matrix)
        N = len(matrix[0])
        dp = [[0] * N for _ in range(M)]
        for n in range(N):
            dp[0][n] = int(matrix[0][n])
        for m in range(M):
            dp[m][0] = int(matrix[m][0])
        for i in range(1, M):
            for j in range(1, N):
                if matrix[i][j] == '0':
                    continue
                dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
        L = numpy.max(dp)
        return L * L		
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运行结果

Runtime: 213 ms, faster than 39.94% of Python online submissions for Maximal Square.
Memory Usage: 33.8 MB, less than 6.85% of Python online submissions for Maximal Square.
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原题链接：leetcode.com/problems/ma…

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