[LeetCode] 85. Maximal Rectangle

LeetCode最大矩形题解

最新推荐文章于 2024-06-20 10:20:06 发布

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最新推荐文章于 2024-06-20 10:20:06 发布

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原文链接：http://www.cnblogs.com/chencode/p/maximal-rectangle.html

本文提供了解决LeetCode上最大矩形问题的两种方法：一种是通过遍历二维数组并利用最大直方图面积的方法求解；另一种是采用动态规划的方法。这两种方法都详细解释了如何找到矩阵中最大的全由1构成的矩形。

https://leetcode.com/problems/maximal-rectangle/

思路一：这道题和前面那道 84. Largest Rectangle in Histogram 类似，遍历二维数组，生成一个 heights 数组，然后用 Largest Rectangle in Histogram 一样的方法做。

public class Solution {
    public int maximalRectangle(char[][] matrix) {
        int result = 0;
        if (matrix == null || matrix.length == 0) {
            return result;
        }
        int[] heights = new int[matrix[0].length];
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                if (matrix[i][j] == '0') {
                    heights[j] = 0;
                } else {
                    heights[j] += 1;
                }
            }
            result = Math.max(result, helper(heights));
        }
        return result;
    }
    
    private int helper(int[] heights) {
        int result = 0;
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < heights.length; i++) {
            while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {
                int tmp = stack.pop();
                int height = heights[tmp];
                int width = stack.isEmpty() ? i : i - stack.peek() - 1;
                result = Math.max(result, height * width);
            }
            stack.push(i);
        }
        while (!stack.isEmpty() && heights[stack.peek()] >= 0) {
            int tmp = stack.pop();
            int height = heights[tmp];
            int width = stack.isEmpty() ? heights.length : heights.length - stack.peek() - 1;
            result = Math.max(result, height * width);
        }
        return result;
    }
}