85. Maximal Rectangle

Violence

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class Solution {
    public int maximalRectangle(char[][] matrix) {
        int m = matrix.length, n = matrix[0].length;

        int[][] tallMatrix = new int[m][n];
        for (int j = 0; j < n; j++) {
            tallMatrix[0][j] = matrix[0][j] == '1' ? 1 : 0;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    if (matrix[i - 1][j] == '1') {
                        tallMatrix[i][j] = tallMatrix[i - 1][j] + 1;
                    } else {
                        tallMatrix[i][j] = 1;
                    }
                }
            }
        }

        int maxArea = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (tallMatrix[i][j] == 0) {
                    continue;
                }
                if (tallMatrix[i][j] * n <= maxArea) {
                    continue;
                }

                // 寻找左右侧高度大于等于当前柱子的所有柱子
                int width = 1; // self
                for (int k = j - 1; k >= 0; k--) {
                    if (tallMatrix[i][k] >= tallMatrix[i][j]) {
                        width++;
                    } else {
                        break;
                    }
                }
                for (int k = j + 1; k < n; k++) {
                    if (tallMatrix[i][k] >= tallMatrix[i][j]) {
                        width++;
                    } else {
                        break;
                    }
                }

                maxArea = Math.max(maxArea, width * tallMatrix[i][j]);
            }
        }

        return maxArea;
    }
}

Stack

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class Solution {
    public int maximalRectangle(char[][] matrix) {
        int m = matrix.length, n = matrix[0].length;

        int maxArea = 0;

        int[] heights = new int[n + 2];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    heights[j + 1] += 1;
                } else {
                    heights[j + 1] = 0;
                }
            }

            Stack<Integer> increaseStack = new Stack<>(); // 非严格单调递增栈
            for (int j = 0; j < heights.length; j++) {
                while (!increaseStack.isEmpty() && heights[j] < heights[increaseStack.peek()]) {
                    // 当前元素比栈顶元素值小
                    int height = heights[increaseStack.pop()];
                    int width = j - increaseStack.peek() - 1;
                    maxArea = Math.max(maxArea, height * width);
                }

                increaseStack.push(j);
            }
        }

        return maxArea;
    }
}

References

85. Maximal Rectangle
剑指 Offer II 040. 矩阵中最大的矩形