未运行二进制矩阵码的最大面积矩形

Question

下面是二进制矩阵最大面积的代码。它有一个函数MAH（）来返回直方图的最大面积。方法是将二进制矩阵（2d）分解为一维。然后将MAH（）应用于每个一维数组并找到最大面积。

类解决方案{

    static class Pair {

        int element;
        int index;

        public Pair(int element, int index) {
            this.element = element;
            this.index = index;
        }

    }
    
     static int MAH(int a[]) {
        Stack<Pair> stack = new Stack<>();
        int n = a.length;
        int nsrIndex[] = new int[n];
        int pseudoIndex = n;

        for (int i = n - 1; i >= 0; i--) {

            while (!stack.isEmpty() && stack.peek().element >= a[i]) {
                stack.pop();
            }
            if (i < n) {
                if (!stack.isEmpty()) {
                    nsrIndex[i] = stack.peek().index;
                } else {
                    nsrIndex[i] = pseudoIndex;
                }
            }
//            push the current element onto the stack
            stack.push(new Pair(a[i], i));
        }

        // System.out.println("NSR -> "+ Arrays.toString(nsrIndex));

        //re-initialize stack to find next smaller to left element's index
        stack = new Stack<>();
        int nslIndex[] = new int[a.length];
        pseudoIndex = -1;

        for (int i = 0; i < n; i++) {

            while (!stack.isEmpty() && stack.peek().element >= a[i]) {
                stack.pop();
            }
            if (i < n) {
                if (!stack.isEmpty()) {
                    nslIndex[i] = stack.peek().index;
                } else {
                    nslIndex[i] = pseudoIndex;
                }
            }
//            push the current element onto the stack
            stack.push(new Pair(a[i], i));
        }
        //System.out.println("NSL ->"+Arrays.toString(nslIndex));

        //width[i] = nsrIndex[i] - nslIndex[i]
        int width[] = new int[n];
        int maxArea = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            width[i] = nsrIndex[i] - nslIndex[i] - 1;
            int area = a[i] * width[i]; //area = a[i] * width[i]
            maxArea = Math.max(area, maxArea);
        }
//        System.out.println("Width ->"+Arrays.toString(width));
//        System.out.println("Max Area "+maxArea);
        return maxArea;
    }
    
    
    
    
    public int maximalRectangle(char[][] matrix) {
        
        int n = matrix.length;
        int m = matrix[0].length;
        
        int b[] = new int[m];
        for (int j = 0; j < m; j++) {

            b[j] = matrix[0][j];
        }
        int max = MAH(b);

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (matrix[i][j] == 0) {
                    b[j] = 0;
                } else {
                    b[j] = b[j] + matrix[i][j];
                }
            }
            max = Math.max(max, MAH(b));
        }
        return max;
}}

问题链接：二进制矩阵的最大面积

MAH（）函数是正确的。仅maximalRectangle（char[][]矩阵）函数无法给出结果。有人能解释一下吗？？

这是另一个人发布的maximalRectangle（char[]]matrix）的代码示例，它运行良好

 int[] arr = new int[matrix[0].length]; 
//         int area = 0;
//         for(int i = 0; i<matrix.length; i++){
//             for(int j = 0; j<matrix[0].length; j++){
//                 if(matrix[i][j] == '1'){
//                     arr[j]++;
//                 }else arr[j] = 0;
                
//             }
//             area = Math.max(area, MAH(arr));
//         }

//         return area; 

Answer 1

我相信你不知何故忽略了输入是char 2d数组，所以，代替b[j]=矩阵[0][j]应该是b[j]=矩阵[0][j]-'0'，而不是矩阵[i][j]==0应该是矩阵[i][j]=='0'等。