[Swift]LeetCode221. 最大正方形 | Maximal Square

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Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example:

Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

---

在一个由 0 和 1 组成的二维矩阵内，找到只包含 1 的最大正方形，并返回其面积。

示例:

输入: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

输出: 4

---

404ms

class Solution {
    func maximalSquare(_ matrix: [[Character]]) -> Int {

        guard matrix.count > 0, matrix[0].count > 0 else { return 0 }
        var m = matrix.count, n = matrix[0].count
        var dp = Array(repeating: Array(repeating: 0, count: n+1), count: m+1)
        var res = 0
        for i in 1...m {
            for j in 1...n {
                if matrix[i-1][j-1] == "0" { continue }
               dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1
                res = max(res, dp[i][j])
            }
        }
        return res * res
    }
}

---

408ms

class Solution {
    func maximalSquare(_ matrix: [[Character]]) -> Int {

        guard matrix.count > 0, matrix[0].count > 0 else { return 0 }
        var m = matrix.count, n = matrix[0].count
        var dp =  Array(repeating: 0, count: n+1)
        var res = 0
        var pre = 0
        for i in 0..<m {
            for j in 1...n {
                var t = dp[j]
                if matrix[i][j-1] == "0" { dp[j] = 0 }
                else {
                    dp[j] = min(min(dp[j], dp[j-1]), pre) + 1
                     res = max(res, dp[j])
                }
                pre = t
            }
        }
        return res * res
    }
}

---

416ms

class Solution {
    func maximalSquare(_ matrix: [[Character]]) -> Int {
        guard !matrix.isEmpty else { return 0 }
        
        var maxLength = 0
        var areaArray = Array(repeating: Array(repeating: 0, count: matrix[0].count), count: matrix.count)
        
        for i in 0..<matrix.count {
            for j in 0..<matrix[0].count {
                if i == 0 || j == 0{
                    areaArray[i][j] = matrix[i][j] == "0" ? 0 : 1
                }else {
                    if matrix[i - 1][j] == "0" && matrix[i][j - 1] == "0" && matrix[i - 1][j - 1] == "0" {
                        areaArray[i][j] = matrix[i][j] == "0" ? 0 : 1
                    }else if matrix[i][j] == "1" {
                        areaArray[i][j] = min(min(areaArray[i - 1][j], areaArray[i][j - 1]), areaArray[i - 1][j - 1]) + 1
                    }
                }
                maxLength = max(maxLength, areaArray[i][j])
            } 
        }
        
        return maxLength * maxLength
    }
}

---

420ms

class Solution {
    func maximalSquare(_ matrix: [[Character]]) -> Int {
      if matrix.count == 0 {
        return 0
      }

      let rows = matrix.count, cols = matrix[0].count
      var maxsqlen = 0

      // make a copy of matrix
      var dp = [[Int]](repeating: ([Int](repeating: 0, count: cols + 1)), count: rows + 1)

      for i in 1...rows {
        for j in 1...cols {
          // if previous i and previous j is 1
          if matrix[i - 1][j - 1] == "1" {
            // find the minimum of [i - 1][j], [i - 1][j - 1], [i][j - 1] and add 1
            dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1], dp[i][j - 1]) + 1
            maxsqlen = max(maxsqlen, dp[i][j])
          }
        }
      }

      return maxsqlen * maxsqlen
    }
}

---

436ms

class Solution {
    func maximalSquare(_ matrix: [[Character]]) -> Int {
        var maxCount = 0
        for row in 0..<matrix.count {
            for col in 0..<matrix[0].count {
                guard matrix[row][col] == "1" else { continue }
                var currentCount = 1
                var base = 1
                let startRow = row
                let startCol = col
                var newRow = row + 1
                var newCol = col + 1
                boundaryCheck: while newRow < matrix.count && newCol < matrix[0].count {
                    for nextRow in startRow...newRow {
                        if matrix[nextRow][newCol] != "1" {
                            break boundaryCheck
                        }
                    }

                    for nextCol in startCol...newCol {
                        if matrix[newRow][nextCol] != "1" {
                            break boundaryCheck
                        }
                    }
                    base += 1
                    currentCount = base * base
                    newRow += 1
                    newCol += 1
                }
                maxCount = max(maxCount, currentCount)
            }
        }
        return maxCount
    }    
}