85. Maximal Rectangle

最新推荐文章于 2021-10-17 09:55:57 发布

原创最新推荐文章于 2021-10-17 09:55:57 发布 · 217 阅读

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本文介绍了一种高效算法，用于解决二维二进制矩阵中寻找包含全部1的最大矩形问题，并通过将其转换为多个直方图来求解每个直方图中的最大矩形面积。

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

受上一题的启发，可以把数组的每一行为基础，分解成一个个的直方图，参考 Largest Rectangle in Histogram的方法进行求解

 public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int row = matrix.length,col = matrix[0].length;
        int maxarea = 0;
        int[][] area = new int[row][col];
        for(int i = 0;i < row;i++){
            for(int j = 0; j < col;j++){
                 if(i == 0){
                     area[i][j] = matrix[i][j]-'0';
                     continue;
                 }
                 if(matrix[i][j] == '1'){
                     area[i][j] = 1+area[i-1][j];
                 }
            }
         }
         for(int i = 0;i < row ; i++){
             maxarea = Math.max(maxarea,maxR(area[i]));
         }
         return maxarea;
    }
    private int maxR(int[] nums){
        LinkedList<Integer> index = new LinkedList<Integer>();
        LinkedList<Integer> height = new LinkedList<Integer>();
        int maxarea = 0;
        for(int i = 0;i < nums.length;i++){
            if(index.isEmpty() || nums[i] >= height.peek()){
                index.push(i);
                height.push(nums[i]);
            }
            else{
                int j = 0;
                while(!index.isEmpty() && nums[i] < height.peek()){
                    j = index.peek();
                    int tmp = (i-index.pop())*height.pop();
                    maxarea = Math.max(tmp,maxarea);
                }
                index.push(j);
                height.push(nums[i]);
            }
        }
        while(!index.isEmpty()){
            int tmp = (nums.length-index.pop())*height.pop();
            maxarea = Math.max(tmp,maxarea);
        }
        return maxarea;
    }