1139. The greatest 1 Formation of the border One question daily

topic：

1139. The greatest 1 Formation of the border.md

Thought：

quiz6 Simple version，Prefix and求解。It should be noted that the upper left corner is0 Special circumstances are not considered，
Need to verify # superior Left Down right Four -edge 1 The number of individuals d

Code：

错误Code

class Solution:
    def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
        matrix_heng = copy.deepcopy(grid)
        matrix_su = copy.deepcopy(grid)
        ans = 0
        if len(grid) == 1:
            if 1 in grid[0]:
                return 1
            else:return 0
        for ind, i in enumerate(grid):
            for index in range(len(i)):
                # Each line adds
                if index >= 1:
                    matrix_heng[ind][index] += matrix_heng[ind][index-1]
                # Add each column
                if ind >= 1:
                    matrix_su[ind][index] += matrix_su[ind-1][index]
        print(matrix_heng, matrix_su)
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                minus = min(matrix_heng[i][j], matrix_su[i][j])
                try:
                    if grid[i-minus+1][j-minus+1] == 1 and i >= minus-1 and j >= minus-1:
                        ans = max(ans, minus ** 2)
                except:
                    pass
        return ans

class Solution:
    def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        rs = [list(accumulate(row, initial=0)) for row in grid]  # 每行的Prefix and
        cs = [list(accumulate(col, initial=0)) for col in zip(*grid)]  # 每列的Prefix and
        for d in range(min(m, n), 0, -1):  # From large to small enumeration square edges d
            for i in range(m - d + 1):
                for j in range(n - d + 1):  # 枚举正方形Leftsuperior角坐标 (i,j)
                    # superior Left Down right Four -edge 1 The number of individuals d
                    if rs[i][j + d] - rs[i][j] == d and \
                       cs[j][i + d] - cs[j][i] == d and \
                       rs[i + d - 1][j + d] - rs[i + d - 1][j] == d and \
                       cs[j + d - 1][i + d] - cs[j + d - 1][i] == d:
                        return d * d
        return 0