leetcode221 最大正方形

芬加达2026-01-27 14:51

java 复制代码

class Solution {
    public int maximalSquare(char[][] matrix) {
        int row = matrix.length;
        int col = matrix[0].length;
        int[][] pref = new int[row + 1][col + 1];
        for (int i = 1; i < row + 1; i++) {
            for (int j = 1; j < col + 1; j++) {
                pref[i][j] = matrix[i - 1][j - 1] - '0' + pref[i - 1][j] + pref[i][j - 1] - pref[i - 1][j - 1];
            }
        }

        int maxSide = 0;

        for (int i = 1; i < row + 1; i++) {
            for (int j = 1; j < col + 1; j++) {
                for (int k = maxSide + 1; i - k >= 0 && j - k >= 0; k++) {
                    int sum = pref[i][j] - pref[i - k][j] - pref[i][j - k] + pref[i - k][j - k];
                    if (sum == k * k)
                        maxSide = k;

                }
            }
        }

        return maxSide * maxSide;

    }
}

在力扣中，这道题的最优解是动态规划（ $O(MN)$ 时间和 $O(N)$ 空间），但前缀和方法对于解决"任意子矩形求和"类问题是通用的大杀器，非常值得掌握。

java 复制代码

public int maximalSquare(char[][] matrix) {
    int rows = matrix.length, cols = matrix[0].length;
    int[][] dp = new int[rows + 1][cols + 1];
    int maxSide = 0;

    for (int i = 1; i <= rows; i++) {
        for (int j = 1; j <= cols; j++) {
            if (matrix[i-1][j-1] == '1') {
                // 状态转移方程
                dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;
                maxSide = Math.max(maxSide, dp[i][j]);
            }
        }
    }
    return maxSide * maxSide;
}