BFS/DFS处理二维矩阵抽象问题例题

Posted 2021-08-13 HardyDragon_CC

这是自己搜集的有关bfs和dfs关于二维01数组的例题

• BFS
• • 烂橘子
  • 01矩阵
  • 图像渲染
• DFS
• • 岛屿数量
  • 岛屿最大面积

BFS

烂橘子

这里队列存储数组下标的方式和其他的不同

class Solution {
    int[] dr = new int[]{-1, 0, 1, 0};
    int[] dc = new int[]{0, -1, 0, 1};

    public int orangesRotting(int[][] grid) {
        int R = grid.length, C = grid[0].length;
        Queue<Integer> queue = new ArrayDeque<Integer>();
        Map<Integer, Integer> depth = new HashMap<Integer, Integer>();
        for (int r = 0; r < R; ++r) {
            for (int c = 0; c < C; ++c) {
                if (grid[r][c] == 2) {
                    // 代替 int[] 队列元素 通过除、取余获取x，y的值
                    int code = r * C + c;
                    queue.add(code);
                    depth.put(code, 0);
                }
            }
        }
        int ans = 0;
        while (!queue.isEmpty()) {
            int code = queue.remove();
            int r = code / C, c = code % C;
            for (int k = 0; k < 4; ++k) {
                int nr = r + dr[k];
                int nc = c + dc[k];
                if (0 <= nr && nr < R && 0 <= nc && nc < C && grid[nr][nc] == 1) {
                    grid[nr][nc] = 2;
                    int ncode = nr * C + nc;
                    queue.add(ncode);
                    // map 存放表示第几层搜索，即返回最后烂掉的橘子需要几分钟
                    depth.put(ncode, depth.get(code) + 1);
                    ans = depth.get(ncode);
                }
            }
        }
        for (int[] row: grid) {
            for (int v: row) {
                if (v == 1) {
                    return -1;
                }
            }
        }
        return ans;
    }
}

01矩阵

这里的方向移动和上面的不同

class Solution {
public int[][] updateMatrix(int[][] matrix) {
        Queue<int[]> queue = new LinkedList<>();
        int[] directions = {-1, 0, 1, 0, -1};

        for (int row = 0; row < matrix.length; row++) {
            for (int col = 0; col < matrix[0].length; col++) {
                if (matrix[row][col] == 0) {
                    queue.offer(new int[] {row, col});
                } else {
                    //标记非零元素为负，和遍历后设定的正数距离加以区分
                    matrix[row][col] = -1;
                }
            }
        }

        int step = 1;
        while (!queue.isEmpty()) {
            //对当前队列中所有零元素遍历，所有元素向四周走一步
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                //获取队列中的元素位置
                int[] cur = queue.poll();
                //向四个方向依次走一步
                for (int j = 0; j < directions.length - 1; j++) {
                    int x = cur[0] + directions[j];
                    int y = cur[1] + directions[j + 1];
                    //如果超出矩阵范围，或者遇见零元素及设置过距离step的元素则跳过，只对未遍历到的-1操作
                    if (x < 0 || x >= matrix.length || y < 0 || y >= matrix[0].length || matrix[x][y] >= 0) {
                        continue;
                    }
                    matrix[x][y] = step;
                    queue.offer(new int[] {x, y});
                }
            }
            //下次遍历到的-1元素相比前一次距离step加1
            step++;
        }
        return matrix;
    }
}

图像渲染

class Solution {
    int[] dx = {1,0,0,-1};
    int[] dy = {0,1,-1,0};
    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        int oldColor = image[sr][sc];
        // 新老颜色相同不变
        if(oldColor == newColor){
            return image;
        }

        int n = image.length;
        int m = image[0].length;
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(new int[]{sr,sc});
        image[sr][sc] = newColor;

        while(!queue.isEmpty()){
            int[] cell = queue.poll();
            int x = cell[0];
            int y = cell[1];
            // 向上下左右走
            for(int i=0;i<4;i++){
                int mx = x + dx[i];
                int my = y + dy[i];
                // 需要不越界且颜色等于老颜色才入队
                if(mx >=0 && mx < n && my>=0 && my < m && image[mx][my] == oldColor){
                    queue.offer(new int[]{mx,my});
                    image[mx][my] = newColor;
                }
            }
        }

        return image;
    }
}

DFS

岛屿数量

class Solution {
    void dfs(char[][] grid, int r, int c) {
        int nr = grid.length;
        int nc = grid[0].length;

        if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
            return;
        }

        grid[r][c] = '0';
        // 上下左右递归dfs搜索
        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    }

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int nr = grid.length;
        int nc = grid[0].length;
        int num_islands = 0;
        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    ++num_islands;
                    dfs(grid, r, c);
                }
            }
        }

        return num_islands;
    }
}

岛屿最大面积

class Solution {
    int dfs(int[][] grid, int r, int c) {
        int nr = grid.length;
        int nc = grid[0].length;

        if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == 0) {
            return 0;
        }

        grid[r][c] = 0;
        int l,right,t,bd
        // 上下左右递归dfs搜索
        l = dfs(grid, r - 1, c);
        right = dfs(grid, r + 1, c);
        b = dfs(grid, r, c - 1);
        t = dfs(grid, r, c + 1);

        return 1 + l + right + t + b; 
    }

    public int maxAreaOfIsland(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int nr = grid.length;
        int nc = grid[0].length;
        int size = 0;
        int res = 0;
        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == 1) {
                    size = dfs(grid, r, c);
                    res = Math.max(res,size);
                }
            }
        }

        return res;
    }
}