[leetcode] 221. Maximal Square @ python

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原题

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:

Input:

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

Output: 4

解法

参考: 花花酱 LeetCode 221. Maximal Square
动态规划. base case是当matrix长度为0时, 返回0. 构造dp数组, 将matrix里的字符串转化为数字, dp[i][j]表示以matrix[i][j]为右下角能构成的最大正方形的边长, 我们从第1行第1列开始, 求dp[i][j].

dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1

然后找出二维数组dp中最大的值就是最大正方形的边, 取平方即可.
Time: O(mn)
Space: O(mn)

代码

class Solution(object):
    def maximalSquare(self, matrix):
        """
        :type matrix: List[List[str]]
        :rtype: int
        """
        # base case
        if not matrix: return 0
        res = 0
        row, col = len(matrix), len(matrix[0])
        dp = [[1 if matrix[i][j] == '1' else 0 for j in range(col)] for i in range(row)]
        for i in range(1, row):
            for j in range(1, col):
                if matrix[i][j] == '1':
                    dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1
                else:
                    dp[i][j] = 0
                    
        for row in dp:
            res = max(res, max(row))
        return res*res