Maximal Rectangle_var maximalrectangle = function(matrix) {}-优快云博客

本文介绍了一种求解二维矩阵中最大全1矩形面积的算法。通过构建高度矩阵并运用单调栈技巧来高效地计算出最大矩形的面积。文章详细展示了算法实现过程，并附带完整的Java代码。

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public class Solution {
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null) {
            throw new IllegalArgumentException("");
        } 
        if (matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int[][] heights = new int[matrix.length][matrix[0].length];
        for (int i = 0; i < heights.length; i++) {
            for (int j = 0; j < heights[0].length; j++) {
                if (i == 0) {
                    heights[i][j] = matrix[i][j] == '1' ? 1 : 0;
                } else {
                    heights[i][j] = matrix[i][j] == '1' ? 1 : 0;
                    if (heights[i][j] != 0) {
                        heights[i][j] = heights[i][j] + heights[i - 1][j];
                    }
                }
            }
        }
        int maxArea = 0;
        int row = heights.length, column = heights[0].length;
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < column; j++) {
                if (stack.isEmpty()) {
                    stack.push(j);
                } else {
                    while (!stack.isEmpty() && heights[i][j] < heights[i][stack.peek()]) {
                        int last_height = heights[i][stack.pop()];
                        if (!stack.isEmpty()) {
                            maxArea = Math.max(maxArea, (j - 1 - stack.peek())*last_height);
                        } else {
                            maxArea = Math.max(maxArea, (j)*last_height);
                        }
                    }
                    stack.push(j);
                }
            }
            while (!stack.isEmpty()) {
                int last_height = heights[i][stack.pop()];
                if (!stack.isEmpty()) {
                    maxArea = Math.max(maxArea, (column - 1 - stack.peek())*last_height);
                } else {
                    maxArea = Math.max(maxArea, (column)*last_height);
                }
            }
        }
        return maxArea;
    }
}