题目：原题链接（中等）
标签：动态规划、数组
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N 2 ) | O ( N 2 ) | 172ms (96.43%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一（动态规划）：
class Solution:
def findSquare(self, matrix: List[List[int]]) -> List[int]:
n = len(matrix)
dp = [[(0, 0)] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(n - 1, -1, -1):
if matrix[i][j] == 0:
x0 = dp[i][j + 1][0] + 1 if j < n - 1 else 1
x1 = dp[i + 1][j][1] + 1 if i < n - 1 else 1
dp[i][j] = (x0, x1)
# for row in dp:
# print(row)
ans_idx, ans_val = (0, 0), 0
for i in range(n):
for j in range(n):
if matrix[i][j] == 0:
size = min(dp[i][j])
if size > ans_val:
for k in range(1, size):
if min(dp[i + k][j + k]) < size - k:
break
else:
ans_idx, ans_val = (i, j), size
return [ans_idx[0], ans_idx[1], ans_val] if ans_val > 0 else []
