使用bfs具有相同值的单元格数

Question

 int[][] matrix = {

  {2 , 3, 4, 10, 12},
  {20 , 30, 14, 11, 13},
  {29 , 39, 40, 12, 24},
  {40 , 39, 39, 15, 35},
  {100 ,23, 24, 60, 80}
  }; 

这应该返回3，因为如果我将从单元格（2,1）开始，我将通过上下左右移动得到39，39，39，我的方法看起来像find_cells(int[][]矩阵，int行，int col)，其中row和col是起点。不要使用任何帮助器方法。我得到1可能是因为我将邻居标记为true，下次当我试图访问它们时，它会跳过它们。很抱歉缩进了。

     public int find_cells(int[][] matrix, int row, int col){
         //invalid row and col
        if (row < 0 | col < 0 | row > matrix.length | col > matrix[0].length)
              return -1;


        // check if cell is already visited.
        boolean[][] visited = new boolean[matrix.length][matrix[0].length];

        //left and right neighbours
       int[] lr_neighbour = {0, 0, -1,1};
       //top and bottom neighbours
       int[] tb_neighbour = {1, -1, 0, 0};

       //number of same cells
       int modified = 0;

      //queue
       Queue<Integer> queue = new LinkedList<>();
       queue.add(matrix[row][col]);
      //mark current cell as visited.
       visited[row][col] = true;

      //current pixel at (row,col)
      int current_cell = matrix[row][col];

     while (!queue.isEmpty()){
        queue.remove();
        for (int index = 0;index < 4;index++){
            row = row+tb_neighbour[index];
            col = col+lr_neighbour[index];
            if (row < 0 || col < 0 || row >= matrix.length || col >= matrix[0].length || visited[row][col])
                continue;
            if (current_cell == matrix[row][col]){
                //mark all other valid cells as visited.
                queue.add(matrix[row][col]);
                modified++;
            }
            visited[row][col] = true;
        }
    }

       return modified;
}

Answer 1

看看这个街区：

for (int index = 0;index < 4;index++){
            row = row+tb_neighbour[index];
            col = col+lr_neighbour[index];

让行=2行=1....

然后在第一次迭代时，u使其:row=2+1=3，col=1+0=1...现在您的行和col值改变了…

public int find_cells(int[][] matrix, int row, int col){
        //invalid row and col
        if (row < 0 | col < 0 | row > matrix.length | col > matrix[0].length)
            return -1;


        // check if cell is already visited.
        boolean[][] visited = new boolean[matrix.length][matrix[0].length];

        //left and right neighbours
        int[] lr_neighbour = {0, 0, -1,1};
        //top and bottom neighbours
        int[] tb_neighbour = {1, -1, 0, 0};

        //number of same cells
        int modified = 0;

        //queue
        Queue<Integer> queue = new LinkedList<>();
        queue.add(row);
        queue.add(col);
        //mark current cell as visited.
        visited[row][col] = true;

        //current pixel at (row,col)
        int current_cell = matrix[row][col];
        int x,y;
        while (!queue.isEmpty()){
            row = queue.remove();
            col = queue.remove();
            for (int index = 0;index < 4;index++){
                x = row+tb_neighbour[index];
                y = col+lr_neighbour[index];
                if (x < 0 || y < 0 || x >= matrix.length || y >= matrix[0].length || visited[x][y])
                    continue;
                if (current_cell == matrix[x][y]){
                    //mark all other valid cells as visited.
                    queue.add(x);
                    queue.add(y);
                    modified++;
                }
                visited[x][y] = true;
            }
        }

        return modified;
    }