LeetCode Maximal Rectangle(dp)



Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

题意：给出一个由字符0和1组成的二维字符数组，找到最大的矩阵，其所有字符为1

思路：动态规划，用left[n]表示所在行的第n个元素的左边界为1的下标，right[n]表示所在行的第n个元素的右边界的下标，height[n]表示所在行的第n个位置连续1的个数，即高度

         状态转移方程为

         left[n] = max{left[n], cur_left}

         right[n] = min{right[n], cur_right}

         height[n] = height[n] + 1 if matrix[row][col] = 1

  代码如下：

class Solution {    public int maximalRectangle(char[][] matrix) {        int row = matrix.length;        if (row == 0) return 0;        int col = matrix[0].length;        int[] left = new int[col];        int[] right = new int[col];        int[] height = new int[col];        Arrays.fill(right, col);        int res = 0;        for (int i = 0; i < row; i++)        {            for (int j = 0; j < col; j++)            {                if (matrix[i][j] == '1') height[j]++;                else height[j] = 0;            }            int cur_left = 0;            for (int j = 0; j < col; j++)            {                if (matrix[i][j] == '1') left[j] = Math.max(left[j], cur_left);                else {                    left[j] = 0;                    cur_left = j + 1;                }            }            int cur_right = col;            for (int j = col - 1; j >= 0; j--)            {                if (matrix[i][j] == '1') right[j] = Math.min(right[j], cur_right);                else {                    right[j] = col;                    cur_right = j;                }            }            for (int j = 0; j < col; j++)            {                res = Math.max(res, (right[j] - left[j]) * height[j]);            }        }        return res;    }}