LeetCode--85--hard--MaximalRectangle

package com.app.main.LeetCode.dynamic;

/**
 * 85
 *
 * hard
 *
 * https://leetcode.com/problems/maximal-rectangle/
 *
 * Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
 *
 * Example:
 *
 * Input:
 * [
 *   ["1","0","1","0","0"],
 *   ["1","0","1","1","1"],
 *   ["1","1","1","1","1"],
 *   ["1","0","0","1","0"]
 * ]
 * Output: 6
 *
 *
 * Created with IDEA
 * author:Dingsheng Huang
 * Date:2019/12/18
 * Time:下午3:20
 */
public class MaximalRectangle {

    int max = 0;

    int[][] matrix2;

    public int maximalRectangle(char[][] matrix) {

        int x = matrix[0].length;
        int y = matrix.length;

        matrix2 = new int[y][x];
        for (int i = 0; i < x; i++) {
            matrix2[0][i] = matrix[0][i] == '1' ? 1 : 0;
            for (int j = 1; j < y; j++) {
                matrix2[j][i] = matrix[j][i] == '1' ? matrix2[j - 1][i] + 1 : matrix2[j - 1][i];
            }
        }
        // divide y
        for (int i = 1; i <= y; i++) {
            process(matrix, i);
        }
        return max;
    }

    private void process(char[][] matrix, int y) {
        for (int i = 0; i < matrix.length; i++) {
            if (i + y  <= matrix.length) {
                help(matrix, i, i + y - 1);
            }
        }
    }

    private void help(char[][] matrix, int y1, int y2) {

        int pre = 0;
        int currMax = 0;
        for (int i = 0; i < matrix[0].length; i++) {
            if (y1 == 0) {
                if ((matrix2[y2][i]) == (y2 - y1 + 1)) {
                    currMax = Math.max(currMax, pre + 1);
                    pre++;
                } else {
                    pre = 0;
                }
            } else {
                if ((matrix2[y2][i] - matrix2[y1 - 1][i]) == (y2 - y1 + 1)) {
                    currMax = Math.max(currMax, pre + 1);
                    pre++;
                } else {
                    pre = 0;
                }
            }

        }
        max = Math.max(currMax * (y2 - y1 + 1), max);
    }
}