[LeetCode][85]maximal-rectangle

佚名 / 2023-08-23 / 原文

Content

Given a rows x cols binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

Example 1:

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 6
Explanation: The maximal rectangle is shown in the above picture.

Example 2:

Input: matrix = [["0"]]
Output: 0

Example 3:

Input: matrix = [["1"]]
Output: 1

Constraints:

rows == matrix.length
cols == matrix[i].length
1 <= row, cols <= 200
matrix[i][j] is '0' or '1'.

Solution

1. 暴力解法

Java

class Solution {
    static final char ONE = '1';
    static final char ZERO = '0';
    public int maximalRectangle(char[][] matrix) {
        // rows == matrix.length
        // cols == matrix[i].length
        // 1 <= row, cols <= 200
        // matrix[i][j] is '0' or '1'.
        int m = matrix.length;
        int n = matrix[0].length;
        int max = 0;
        // dp[i][j][0]表示以matrix[i][j]为右端点的水平最长连续'1'的长度
        // dp[i][j][1]表示以matrix[i][j]为下端点的垂直最长连续'1'的长度
        int[][][] dp = new int[m][n][2];
        // 初始化xy
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = new int[]{0, 0};
            }
        }
        if (matrix[0][0] == ONE) {
            dp[0][0][0] = 1;
            dp[0][0][1] = 1;
            max = 1;
        }
        for (int j = 1; j < n; j++) {
            if (matrix[0][j] == ONE) {
                dp[0][j][0] = dp[0][j - 1][0] + 1;
                dp[0][j][1] = 1;
                max = Math.max(max, dp[0][j][0]);
            }
        }
        for (int i = 1; i < m; i++) {
            if (matrix[i][0] == ONE) {
                dp[i][0][0] = 1;
                dp[i][0][1] = dp[i - 1][0][1] + 1;
                max = Math.max(max, dp[i][0][1]);
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == ONE) {
                    dp[i][j][0] = dp[i][j - 1][0] + 1;
                    dp[i][j][1] = dp[i - 1][j][1] + 1;
                    int x = Integer.MAX_VALUE;
                    int y = 0;
                    int i0 = i;
                    while (i - i0 + 1 <= dp[i][j][1]) {
                        x = Math.min(x, dp[i0][j][0]);
                        y++;
                        max = Math.max(max, x * y);
                        i0--;
                    }
                }
            }
        }
        return max;
    }
}