题解 | #最大矩形#

做这题之前建议先去做力扣 84. 柱状图中最大的矩形，这题实际上是遍历每一行，以每一行为底形成一个柱状图， 柱状图中最大的矩形，然后取得最大值。

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param matrix int整型二维数组
     * @return int整型
     */
     public int maximalRectangle(int[][] matrix) {
        int n = matrix.length, m = matrix[0].length;
        int[] height = new int[m];
        int maxArea=0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (matrix[i][j] == 1) height[j]++;
                else height[j] = 0;
            }
            maxArea=Math.max(maxArea,getMaxArea(height));
        }
        return maxArea;
    }

    public int getMaxArea(int[] height) {
        Deque<Integer> stack = new LinkedList<>();
        int[] newHeight = new int[height.length + 2];
        newHeight[0] = 0;
        newHeight[newHeight.length - 1] = 0;
        System.arraycopy(height, 0, newHeight, 1, height.length);
        stack.push(0);
        int maxArea=0;
        for (int i = 1; i < newHeight.length; i++) {
            if(newHeight[i]>=newHeight[stack.peek()]){
                stack.push(i);
            }else{
                while (!stack.isEmpty()&&newHeight[i]<newHeight[stack.peek()]){
                    int mid=stack.pop();
                    if(!stack.isEmpty()){
                        int left=stack.peek();
                        int area=(i-left-1)*newHeight[mid];
                        maxArea=Math.max(maxArea,area);
                    }
                }
                stack.push(i);
            }
        }
        return maxArea;
    }
}