1139. 最大的以 1 为边界的正方形

题目：
给你一个由若干 0 和 1 组成的二维网格 grid，请你找出边界全部由 1 组成的最大 正方形 子网格，并返回该子网格中的元素数量。如果不存在，则返回 0。

算法：
方法一:前缀和

func largest1BorderedSquare(grid [][]int) int {
    n, m := len(grid), len(grid[0])

    rs := make([][]int, n)
    for i := range rs {
        rs[i] = make([]int, m + 1)
    }

    cs := make([][]int, m)
    for i := range cs {
        cs[i] = make([]int, n + 1)
    }
    for i := range grid {
        for j := range grid[i] {
            rs[i][j + 1] = rs[i][j] + grid[i][j]
            cs[j][i + 1] = cs[j][i] + grid[i][j]
        } 
    }

    for d := min(m, n); d > 0; d -- {
        for i := 0; i <= n - d; i ++ {
            for j := 0; j <= m - d; j ++ {
                if rs[i][j + d] - rs[i][j] == d &&
                cs[j][i + d] - cs[j][i] == d &&
                rs[i + d - 1][j + d] - rs[i + d - 1][j] == d &&
                cs[j + d - 1][i + d ] - cs[j + d - 1][i] == d {
                    return d * d
                }   
            }
        }
    }

    return 0
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

方法二：前缀和

func largest1BorderedSquare(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  // row,和col方向上，最大连续的1的个数
  row, col := make([][]int, m + 1), make([][]int, m + 1)
  largestEdge := 0
  for i := range col {
    row[i] = make([]int, n + 1)
    col[i] = make([]int, n + 1)
  }

  for i := range grid {
    for j := range grid[i] {
      if grid[i][j] == 1 {
        row[i + 1][j + 1] = row[i][j + 1] + 1
        col[i + 1][j + 1] = col[i + 1][j] + 1
      }
      edge := min(row[i + 1][j + 1], col[i + 1][j + 1])
      for k := edge; k > 0; k -- {
        if j - k + 2 > 0 && row[i + 1][j - k + 2] >= k && i - k + 2 > 0 && col[i - k + 2][j + 1] >= k {
          if k > largestEdge {
            largestEdge = k
            break
          }
        }
      }
    }
  }

  return largestEdge * largestEdge
}

func min(a, b int) int {
  if a < b {
    return a
  }
  return b
}