题目描述

• 其实是84. 柱状图中最大的矩形的兄弟题目，理解成多个84题，对结果取max即可。
  
  

思路 && 代码

• 一行抽象出一个【柱状图】，分别套到84题的函数里即可
• 时空复杂度：O($n^2$)、O(n)

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0) {
            return 0;
        }

        // 看成【对每层，进行柱状图最大面积判断】即可（相当于固定底）
        int max = 0;
        int[] heights = new int[matrix[0].length];
        for(int i = 0; i < matrix.length; i++) {
            for(int j = 0; j < matrix[0].length; j++) {
                if(matrix[i][j] == '1') {
                    heights[j]++;
                }
                else {
                    heights[j] = 0;
                }
            }
            max = Math.max(max, largestRectangleArea(heights));
        }
        return max;
    }

    // 84. 求柱状图最大矩阵面积
    public int largestRectangleArea(int[] heights) {
        int res = 0;
        Deque<Integer> stack = new ArrayDeque<>();
        int[] newHeights = new int[heights.length + 2];
        for(int i = 1; i < heights.length + 1; i++) {
            newHeights[i] = heights[i - 1];
        }
        for(int i = 0; i < newHeights.length; i++) {
            while(!stack.isEmpty() && newHeights[stack.peek()] > newHeights[i]) {
                int index = stack.pop();
                int l = stack.peek();
                int r = i;
                res = Math.max(res, (r - l - 1) * newHeights[index]);
            }
            stack.push(i);
        }
        return res;
    }
}

二刷

• 思路还是记得的

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0) return 0;

        int[] heights = new int[matrix[0].length];
        int res = 0;
        // 逐行转换
        for(int i = 0; i < matrix.length; i++) {
            // 当前行的逐列维护
            for(int j = 0; j < matrix[0].length; j++) {
                if(matrix[i][j] == '1') {
                    heights[j]++;
                }
                else {
                    heights[j] = 0;
                }
            }
            res = Math.max(res, largestRectangleArea(heights));
        }
        return res;
    }
    
    public int largestRectangleArea(int[] heights) {
        int[] newHeights = new int[heights.length + 2];
        for(int i = 1; i <= heights.length; i++) {
            newHeights[i] = heights[i - 1];
        }

        Deque<Integer> stack = new ArrayDeque<>();
        int max = 0;
        for(int i = 0; i < newHeights.length; i++) {
            while(!stack.isEmpty() && newHeights[i] < newHeights[stack.peek()]) {
                int now = stack.poll();
                int left = stack.peek(); 
                int right = i; 
                max = Math.max(max, (right - left - 1) * newHeights[now]);
            }
            stack.push(i);
        }
        return max;
    }
}