lc 最大正方形

链接:https://leetcode-cn.com/problems/maximal-square/

代码:

class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        m = len(matrix)
        if m == 0:
            return 0
        res = -1
        n = len(matrix[0])
        dp = [[0] * n for _ in range(m)]
        if n == 0:
            return 0
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == '1':
                    if i == 0 or j == 0:
                        dp[i][j] = 1
                    else:
                        dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1
                res = max(res, dp[i][j])
        return res*res
View Code

思路:dp[i,j]表示以 matrix[i,j]为右下角的最大的正方形的边长。dp[i,j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1,是由左,上,左上转移来的,一定要记住 dp 是又同等性质不同规模的子问题转移过来,通过记录子问题状态来达到节省时间的目的(不用每到一个点就重复算子问题了)

原文地址:https://www.cnblogs.com/FriskyPuppy/p/12953846.html