最大正方形
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 最大正方形
# @param matrix char字符型二维数组
# @return int整型
#
# @param matrix char字符型二维数组
# @return int整型
#
class Solution:
def solve(self, matrix: List[List[str]]) -> int:
# write code here
# 方法2 找重合矩阵的最短边， 往右下角遍历， 每次找 dp[i][j-1], dp[i-1][j-1], dp[i-1][j]最短边 +1， 这个1 代表当前为1的元素， 变成加1
max_len = 0
if len(matrix) == 0:
return 0
for j in range(len(matrix)):
print(matrix[j])
rows = len(matrix)
cols = len(matrix[0])
dp = [[0 for i in range(cols+1) ] for j in range(rows+1) ]
for r in range(1, rows+1):
for c in range(1, cols+1):
if matrix[r-1][c-1] == '1':
min_val = min(dp[r-1][c], dp[r][c-1])
dp[r][c] = min(dp[r-1][c-1], min_val)+1
max_len = max(max_len, dp[r][c])
# print("dp row",dp[r], 'r:', r)
for j in range(len(dp)):
print("dp: ", dp[j], j)
# print(dp)
return max_len*max_len
# import math
# class Solution:
# def detect_rectangle(self, r, c, m, dp):
# import math
# length = int(math.sqrt(dp[r - 1][c]))
# # print('length:', length, 'r, c',r,c)
# l_cnt = 0
# for i in range(length + 1):
# flag = (
# r < len(m)
# and c + i < len(m[0])
# and r - 1 > 0
# and m[r][c + i] == "1"
# and m[r - 1][c + i] == "1"
# )
# if not flag:
# # return False
# break
# l_cnt += 1
# area = l_cnt * l_cnt
# return area
# def solve(self, matrix: List[List[str]]) -> int:
# # write code here
# # [[1,0,1,0,0],
# # [1,0,1,1,1],
# # [1,1,1,1,1],
# # [1,0,0,1,0]]
# # 问题拆解： 最大， 正方形
# # 最大： 贪心算法/ DP
# # 正方形： 怎么定义正方形， 怎么检查正方形
# # 最小正方形： 1个元素1
# # 遍历每一行怎么检查正方形：
# # 1. dp矩阵存放以当前元素为左下角的正方形最大面积
# # 2. 每次遍历到 m[r][c], 就找 dp[r-1][c]在相邻正方形最大的边长
# # 3. 然后 m[r][c] 往m[r][c+1]， m[r-1][c+1] 方向检查 最大的边长次数的看是否能形成更大的正方形
# max_square = 0
# if len(matrix) == 0:
# return 0
# rows = len(matrix)
# cols = len(matrix[0])
# dp = [[0] * cols] * rows
# for r in range(rows):
# for c in range(cols):
# if matrix[r][c] == "0":
# dp[r][c] = 0
# else:
# if r == 0:
# dp[r][c] = 1
# else:
# area = self.detect_rectangle(r, c, matrix, dp)
# area = max(area, 1)
# dp[r][c] = area
# # dp[r][c] = (int(math.sqrt(dp[r - 1][c])) + 1) ** 2
# # else:
# # dp[r][c] = 1
# max_square = max(max_square, dp[r][c])
# print("dp row",dp[r], 'r:', r)
# # print("dp: ", dp)
# return max_square
# # dp: [
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # [1, 0, 1, 1, 0, 4, 1, 0, 0, 0],
# # ]
# [
# [1, 0, 1, 1, 1, 1, 1, 0, 0, 0],
# [1, 0, 1, 1, 1, 0, 0, 0, 0, 0],
# [1, 0, 1, 0, 1, 1, 1, 0, 0, 0],
# [1, 1, 0, 1, 1, 1, 1, 0, 1, 0],
# [1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
# [1, 0, 1, 1, 1, 1, 1, 1, 1, 0],
# [1, 0, 1, 1, 1, 1, 1, 1, 1, 0],
# [1, 0, 1, 1, 1, 1, 1, 1, 1, 0],
# [1, 1, 0, 1, 1, 1, 1, 1, 0, 0],
# [1, 0, 1, 1, 0, 1, 1, 0, 0, 0],
# ]
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