python3 解法, 执行用时: 248 ms, 内存消耗: 28.1 MB, 提交时间: 2022-11-24 11:20:38

class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        if len(matrix) == 0 or len(matrix[0]) == 0:
            return 0
        
        maxSide = 0
        rows, columns = len(matrix), len(matrix[0])
        # dp[i][j] 表示以 (i, j) 为右下角，最大正方形边长
        dp = [[0] * columns for _ in range(rows)]
        for i in range(rows):
            for j in range(columns):
                if matrix[i][j] == '1':
                    if i == 0 or j == 0:
                        dp[i][j] = 1
                    else:
                        dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1
                    maxSide = max(maxSide, dp[i][j])
        
        maxSquare = maxSide * maxSide
        return maxSquare

golang 解法, 执行用时: 8 ms, 内存消耗: 6.5 MB, 提交时间: 2022-11-24 11:17:59

// 动态规划
func maximalSquare(matrix [][]byte) int {
    dp := make([][]int, len(matrix))
    maxSide := 0
    for i := 0; i < len(matrix); i++ {
        dp[i] = make([]int, len(matrix[i]))
        for j := 0; j < len(matrix[i]); j++ {
            dp[i][j] = int(matrix[i][j] - '0')
            if dp[i][j] == 1 {
                maxSide = 1
            }
        }
    }

    for i := 1; i < len(matrix); i++ {
        for j := 1; j < len(matrix[i]); j++ {
            if dp[i][j] == 1 {
                dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1
                if dp[i][j] > maxSide {
                    maxSide = dp[i][j]
                }
            }
        }
    }
    return maxSide * maxSide
}

func min(x, y int) int {
    if x < y {
        return x
    }
    return y
}