最大正方形Python解法

在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。

例：

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximal-square

解析：

动态规划，dp数组，边界情况为1直接填1，否则填0。其他位置，因为是正方形，所以左边和上边的值中短的一条边加上左上角使我们的判断条件，因为我们求的是右下角的位置。知道这个规则就可以填写dp数组了。

class Solution(object):def maximalSquare(self, matrix):""":type matrix: List[List[str]]:rtype: int"""if len(matrix) == 0 or len(matrix[0]) == 0:  # 判空return 0maxSide = 0  # 最长边的长度rows, columns = len(matrix), len(matrix[0])  # 获取长和宽dp = [[0] * columns for _ in range(rows)]  # 初始化dp数组for i in range(rows):  # 遍历for j in range(columns):if matrix[i][j] == '1':  # 有值才有正方形if i == 0 or j == 0:  # 边界情况dp[i][j] = 1else:dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1  # 主要判断方法maxSide = max(maxSide, dp[i][j])  # 去最大值maxSquare = maxSide * maxSide  # 求面积return maxSquare

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