85. 最大矩形

Posted 2021-12-06 潜行前行

85. 最大矩形

给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。

示例 1：

输入：matrix = [[“1”,“0”,“1”,“0”,“0”],[“1”,“0”,“1”,“1”,“1”],[“1”,“1”,“1”,“1”,“1”],[“1”,“0”,“0”,“1”,“0”]]
输出：6
解释：最大矩形如上图所示

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix.length == 0 || matrix[0].length ==0 ) return 0;
        int[][] height = new int[matrix.length][matrix[0].length];
        for(int i = 0;i<matrix.length;i++){
            height[i][0] = matrix[i][0] - '0';
            for(int j = 1;j< matrix[0].length;j++){
                height[i][j] = matrix[i][j] == '1' ?  height[i][j-1] + 1 : 0;
            }
        }
        int max = 0;
        for(int i = 0;i<matrix[0].length;i++){
            int[] maxLeft = new int[matrix.length];
            Stack<Integer> stack = new Stack<>();
            stack.push(0);
            int j = 1;
            for(;j < matrix.length;j++){
                if(height[j][i] > height[j-1][i]){
                    stack.push(j);
                    maxLeft[j] = j;
                }else{
                    int index = 0;
                    while (stack.size() > 0 && height[j][i] <= height[stack.peek()][i]){
                        index = stack.pop();
                        max = Math.max(height[index][i] * (j - maxLeft[index]), max);
                    }
                    maxLeft[j] = stack.empty() ? 0 : maxLeft[index];
                    stack.push(j);
                }
            }
            while (stack.size() > 0) {
                int index = stack.pop();
                max = Math.max(height[index][i] * (j - maxLeft[index]), max);
            }
        }
        return max;
    }
}