题目：原题链接（中等）
标签：动态规划
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( M × N ) | O ( M × N ) | 180ms (56.30%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
dp = [[[0, 0, 0, 0] for _ in range(n)] for _ in range(m)] # 左上右下
for i in range(m):
for j in range(n):
if grid[i][j] == 1:
if i > 0:
dp[i][j][0] = dp[i - 1][j][0] + 1
else:
dp[i][j][0] = 1
if j > 0:
dp[i][j][1] = dp[i][j - 1][1] + 1
else:
dp[i][j][1] = 1
ans = 0
for i in range(m - 1, -1, -1):
for j in range(n - 1, -1, -1):
if grid[i][j] == 1:
if i < m - 1:
dp[i][j][2] = dp[i + 1][j][2] + 1
else:
dp[i][j][2] = 1
if j < n - 1:
dp[i][j][3] = dp[i][j + 1][3] + 1
else:
dp[i][j][3] = 1
v1 = min(dp[i][j][2], dp[i][j][3])
if v1 > ans:
for v2 in range(ans + 1, v1 + 1):
if v2 > ans:
if min(dp[i + v2 - 1][j + v2 - 1][0], dp[i + v2 - 1][j + v2 - 1][1]) >= v2:
ans = v2
return ans * ans
