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221. 最大正方形

题目描述

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 1：

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4

示例 2：

输入：matrix = [["0","1"],["1","0"]]
输出：1

示例 3：

输入：matrix = [["0"]]
输出：0

 

提示：

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 300
  • matrix[i][j] 为 '0' 或 '1'

解法

方法一：动态规划

我们定义 \(dp[i + 1][j + 1]\) 表示以下标 \((i, j)\) 作为正方形右下角的最大正方形边长。答案为所有 \(dp[i + 1][j + 1]\) 中的最大值。

状态转移方程为：

\[ dp[i + 1][j + 1] = \begin{cases} 0 & \textit{if } matrix[i][j] = '0' \\ \min(dp[i][j], dp[i][j + 1], dp[i + 1][j]) + 1 & \textit{if } matrix[i][j] = '1' \end{cases} \]

时间复杂度 \(O(m\times n)\)，空间复杂度 \(O(m\times n)\)。其中 \(m\) 和 \(n\) 分别是矩阵的行数和列数。

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class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        m, n = len(matrix), len(matrix[0])
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        mx = 0
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == '1':
                    dp[i + 1][j + 1] = min(dp[i][j + 1], dp[i + 1][j], dp[i][j]) + 1
                    mx = max(mx, dp[i + 1][j + 1])
        return mx * mx

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class Solution {
    public int maximalSquare(char[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[][] dp = new int[m + 1][n + 1];
        int mx = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == '1') {
                    dp[i + 1][j + 1] = Math.min(Math.min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
                    mx = Math.max(mx, dp[i + 1][j + 1]);
                }
            }
        }
        return mx * mx;
    }
}

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class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        int mx = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == '1') {
                    dp[i + 1][j + 1] = min(min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
                    mx = max(mx, dp[i + 1][j + 1]);
                }
            }
        }
        return mx * mx;
    }
};

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func maximalSquare(matrix [][]byte) int {
    m, n := len(matrix), len(matrix[0])
    dp := make([][]int, m+1)
    for i := 0; i <= m; i++ {
        dp[i] = make([]int, n+1)
    }
    mx := 0
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if matrix[i][j] == '1' {
                dp[i+1][j+1] = min(min(dp[i][j+1], dp[i+1][j]), dp[i][j]) + 1
                mx = max(mx, dp[i+1][j+1])
            }
        }
    }
    return mx * mx
}

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public class Solution {
    public int MaximalSquare(char[][] matrix) {
        int m = matrix.Length, n = matrix[0].Length;
        var dp = new int[m + 1, n + 1];
        int mx = 0;
        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (matrix[i][j] == '1')
                {
                    dp[i + 1, j + 1] = Math.Min(Math.Min(dp[i, j + 1], dp[i + 1, j]), dp[i, j]) + 1;
                    mx = Math.Max(mx, dp[i + 1, j + 1]);
                }
            }
        }
        return mx * mx;
    }
}

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