LeetCode —— dfs和bfs

797. 所有可能的路径

给你一个有 n 个节点的 有向无环图(DAG),请你找出所有从节点 0 到节点 n-1 的路径并输出(不要求按特定顺序)。 graph[i] 是一个从节点 i 可以访问的所有节点的列表(即从节点 i 到节点 graph[i][j]存在一条有向边)。

示例1:输入:graph = [[1,2],[3],[3],[]]         输出:[[0,1,3],[0,2,3]] 

示例2:输入:graph = [[4,3,1],[3,2,4],[3],[4],[]]      输出:[[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]

// dfs
class Solution {
    List> list = new ArrayList<>();
    LinkedList path = new LinkedList<>();

    public List> allPathsSourceTarget(int[][] graph) {
        path.add(0);
        dfs(graph, 0);
        return list;
    }

    public void dfs(int[][] graph, int node){
        if(node == graph.length - 1){
            list.add(new ArrayList<>(path));
            return;
        }

        for(int i = 0; i < graph[node].length; i++){
            path.add(graph[node][i]);
            dfs(graph, graph[node][i]);
            path.removeLast();
        }
    }
}

200. 岛屿数量

给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。此外,你可以假设该网格的四条边均被水包围。

示例1:

输入:grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出:1
// dfs (以下三种方式等价)
class Solution {
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == '1'){
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void dfs(char[][] grid, int i, int j){
        if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0'){
            return;
        }
        grid[i][j] = '0';
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}


class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == '1'){
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void dfs(char[][] grid, int i, int j){
        if(grid[i][j] == '0'){
            return;
        }

        grid[i][j] = '0';
        for(int[] pos : positions){
            int newRow = i + pos[0];
            int newColumn = j + pos[1];
            if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length){
                continue;
            }
            dfs(grid, newRow, newColumn);
        }
    }
}


class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    boolean[][] visited;        // 通过判断该位置是否被访问过,而不是将1修改为0

    public int numIslands(char[][] grid) {
        visited = new boolean[grid.length][grid[0].length];
        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == '1' && !visited[i][j]){
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void dfs(char[][] grid, int i, int j){
        if(visited[i][j] || grid[i][j] == '0'){
            return;
        }

        visited[i][j] = true;
        for(int[] pos : positions){
            int newRow = i + pos[0];
            int newColumn = j + pos[1];
            if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length){
                continue;
            }
            dfs(grid, newRow, newColumn);
        }
    }
}
// bfs      以下三种方式等价
class Solution {
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == '1'){
                    count++;
                    bfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void bfs(char[][] grid, int i, int j){
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            i = cur[0];
            j = cur[1];
            if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0'){
                continue;
            }
            grid[i][j] = '0';
            queue.offer(new int[]{i - 1, j});
            queue.offer(new int[]{i + 1, j});
            queue.offer(new int[]{i, j - 1});
            queue.offer(new int[]{i, j + 1});
        }
    }
}


class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == '1'){
                    count++;
                    bfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void bfs(char[][] grid, int i, int j){
        Queue queue = new LinkedList();
        queue.offer(new int[]{i, j});
        grid[i][j] = '0';
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            for(int[] pos : positions){
                int newRow = cur[0] + pos[0];
                int newColumn = cur[1] + pos[1];
                if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length || grid[newRow][newColumn] == '0'){
                    continue;
                }
                queue.offer(new int[]{newRow, newColumn});
                grid[newRow][newColumn] = '0';
            }
        }
    }
}


class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    boolean[][] visited;        // 通过判断该位置是否被访问过,而不是将1修改为0

    public int numIslands(char[][] grid) {
        visited = new boolean[grid.length][grid[0].length];
        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == '1' && !visited[i][j]){
                    count++;
                    bfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void bfs(char[][] grid, int i, int j){
        Queue queue = new LinkedList();
        queue.offer(new int[]{i, j});
        visited[i][j] = true;
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            for(int[] pos : positions){
                int newRow = cur[0] + pos[0];
                int newColumn = cur[1] + pos[1];
                if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length || grid[newRow][newColumn] == '0' || visited[newRow][newColumn]){
                    continue;
                }
                queue.offer(new int[]{newRow, newColumn});
                visited[newRow][newColumn] = true;
            }
        }
    }
}

695. 岛屿的最大面积

给你一个大小为 m x n 的二进制矩阵 grid 。岛屿 是由一些相邻的 1 (代表土地) 构成的组合,这里的「相邻」要求两个 1 必须在 水平或者竖直的四个方向上 相邻。你可以假设 grid 的四个边缘都被 0(代表水)包围着。岛屿的面积是岛上值为 1 的单元格的数目。计算并返回 grid 中最大的岛屿面积。如果没有岛屿,则返回面积为 0 。

LeetCode —— dfs和bfs_第1张图片

输入:grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
输出:6
// dfs
class Solution {
    int area;
    public int maxAreaOfIsland(int[][] grid) {
        int maxArea = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    area = 0;
                    dfs(grid, i, j);
                    maxArea = Math.max(maxArea, area);
                }
            }
        }
        return maxArea;
    }

    public void dfs(int[][] grid, int i, int j){
        if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
            return;
        }
        area++;
        grid[i][j] = 0;
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}
// bfs
class Solution {
    int area;
    public int maxAreaOfIsland(int[][] grid) {
        int maxArea = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    area = 0;
                    bfs(grid, i, j);
                    maxArea = Math.max(maxArea, area);
                }
            }
        }
        return maxArea;
    }

    public void bfs(int[][] grid, int i, int j){
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            i = cur[0];
            j = cur[1];
            if(i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
                area++;
                grid[i][j] = 0;
                queue.offer(new int[]{i - 1, j});
                queue.offer(new int[]{i + 1, j});
                queue.offer(new int[]{i, j - 1});
                queue.offer(new int[]{i, j + 1});
            }
        }
    }
}

463. 岛屿的周长

给定一个 row x col 的二维网格地图 grid ,其中:grid[i][j] = 1 表示陆地, grid[i][j] = 0 表示水域。网格中的格子 水平和垂直 方向相连(对角线方向不相连)。整个网格被水完全包围,但其中恰好有一个岛屿(或者说,一个或多个表示陆地的格子相连组成的岛屿)。岛屿中没有“湖”(“湖” 指水域在岛屿内部且不和岛屿周围的水相连)。格子是边长为 1 的正方形。网格为长方形,且宽度和高度均不超过 100 。计算这个岛屿的周长。

LeetCode —— dfs和bfs_第2张图片

示例:输入:grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]         输出:16

// dfs
class Solution {
    int count = 0;
    public int islandPerimeter(int[][] grid) {
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void dfs(int[][] grid, int i, int j){
        if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
            count++;
            return;
        }
        if(grid[i][j] == 2){
            return;
        }
        grid[i][j] = 2;
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}


class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    boolean[][] visited;
    int count;
    public int islandPerimeter(int[][] grid) {
        visited = new boolean[grid.length][grid[0].length];
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1 && !visited[i][j]){
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void dfs(int[][] grid, int i, int j){
        if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
            count++;
            return;
        }
        if(visited[i][j] && grid[i][j] == 1){
            return;
        }

        visited[i][j] = true;
        for(int[] pos : positions){
            int newRow = i + pos[0];
            int newColumn = j + pos[1];
            dfs(grid, newRow, newColumn);
        }
    }
}
// bfs
class Solution {
    int count = 0;
    public int islandPerimeter(int[][] grid) {
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    bfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void bfs(int[][] grid, int i, int j){
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            i = cur[0];
            j = cur[1];
            if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
                count++;
                continue;
            }
            if(grid[i][j] == 2){
                continue;
            }
            
            grid[i][j] = 2;
            queue.offer(new int[]{i - 1, j});
            queue.offer(new int[]{i + 1, j});
            queue.offer(new int[]{i, j - 1});
            queue.offer(new int[]{i, j + 1});
        }
    }
}


class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    boolean[][] visited;
    int count = 0;
    public int islandPerimeter(int[][] grid) {
        visited = new boolean[grid.length][grid[0].length];
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1 && !visited[i][j]){
                    bfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void bfs(int[][] grid, int i, int j){
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        visited[i][j] = true;
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            for(int[] pos : positions){
                int newRow = cur[0] + pos[0];
                int newColumn = cur[1] + pos[1];
                if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length || grid[newRow][newColumn] == 0){
                    count++;
                    continue;
                }
                if(visited[newRow][newColumn] && grid[newRow][newColumn] == 1){
                    continue;
                }
                queue.offer(new int[]{newRow, newColumn});
                visited[newRow][newColumn] = true;
            }
        }
    }
}

827. 最大人工岛屿

给你一个大小为 n x n 二进制矩阵 grid 。最多 只能将一格 0 变成 1 。返回执行此操作后,grid 中最大的岛屿面积是多少?岛屿 由一组上、下、左、右四个方向相连的 1 形成。

示例:输入: grid = [[1, 0], [0, 1]]         输出: 3

// dfs
class Solution {
    int islandSize;        // 每个岛屿的大小

    public int largestIsland(int[][] grid) {
        int[][] positions = new int[][]{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        int m = grid.length;
        int n = grid[0].length;

        Map map = new HashMap<>();    // key为岛屿编号,value为岛屿大小
        int islandNumber = 2;                           // 岛屿编号从2开始,与0和1进行区分
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    islandSize = 0;
                    dfs(grid, i, j, islandNumber);
                    map.put(islandNumber, islandSize);
                    islandNumber++;
                }
            }
        }

        int result = Integer.MIN_VALUE;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] != 0){
                    continue;
                }
                int curSize = 1;
                Set set = new HashSet<>();
                for(int[] pos : positions){
                    int newRow = i + pos[0];
                    int newColumn = j + pos[1];
                    if(newRow < 0 || newRow >= m || newColumn < 0 || newColumn >= n || grid[newRow][newColumn] == 0){
                        continue;
                    }
                    islandNumber = grid[newRow][newColumn];
                    if(set.contains(islandNumber) || !map.containsKey(islandNumber)){
                        continue;
                    }
                    set.add(islandNumber);
                    curSize += map.get(islandNumber);
                }
                result = Math.max(result, curSize);
            }
        }
        return result == Integer.MIN_VALUE ? m * n : result;
    }

    public void dfs(int[][] grid, int i, int j, int islandNumber){
        if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] != 1){
            return;
        }
        islandSize++;
        grid[i][j] = islandNumber;
        dfs(grid, i - 1, j, islandNumber);
        dfs(grid, i + 1, j, islandNumber);
        dfs(grid, i, j - 1, islandNumber);
        dfs(grid, i, j + 1, islandNumber);
    }
}

1020. 飞地的数量

给你一个大小为 m x n 的二进制矩阵 grid ,其中 0 表示一个海洋单元格、1 表示一个陆地单元格。一次 移动 是指从一个陆地单元格走到另一个相邻(上、下、左、右)的陆地单元格或跨过 grid 的边界。返回网格中 无法 在任意次数的移动中离开网格边界的陆地单元格的数量。

LeetCode —— dfs和bfs_第3张图片

示例:输入:grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]         输出:3

// dfs
class Solution {
    public int numEnclaves(int[][] grid) {
        for(int i = 0; i < grid.length; i++){   
            if(grid[i][0] == 1){      // 左侧
                dfs(grid, i, 0);
            }
            if(grid[i][grid[0].length - 1] == 1){     // 右侧
                dfs(grid, i, grid[0].length - 1);
            }
        }
        for(int j = 1; j < grid[0].length - 1; j++){    
            if(grid[0][j] == 1){      // 上侧
                dfs(grid, 0, j);
            }
            if(grid[grid.length - 1][j] == 1){      // 下侧
                dfs(grid, grid.length - 1, j);
            }
        }

        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    count++;
                }
            }
        }
        return count;
    }

    public void dfs(int[][] grid, int i, int j){
        if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
            return;
        }
        grid[i][j] = 0;
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}
// bfs
class Solution {
    public int numEnclaves(int[][] grid) {
        for(int i = 0; i < grid.length; i++){
            if(grid[i][0] == 1){      // 左侧
                bfs(grid, i, 0);
            }
            if(grid[i][grid[0].length - 1] == 1){     // 右侧
                bfs(grid, i, grid[0].length - 1);
            }
        }
        for(int j = 1; j < grid[0].length - 1; j++){
            if(grid[0][j] == 1){      // 上侧
                bfs(grid, 0, j);
            }
            if(grid[grid.length - 1][j] == 1){      // 下侧
                bfs(grid, grid.length - 1, j);
            }
        }

        int count = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    count++;
                }
            }
        }
        return count;
    }

    public void bfs(int[][] grid, int i, int j){
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            i = cur[0];
            j = cur[1];
            if(i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
                grid[i][j] = 0;
                queue.offer(new int[]{i - 1, j});
                queue.offer(new int[]{i + 1, j});
                queue.offer(new int[]{i, j - 1});
                queue.offer(new int[]{i, j + 1});
            }
        }
    }
}

130. 被围绕的区域

给你一个 m x n 的矩阵 board ,由若干字符 'X' 和 'O' ,找到所有被 'X' 围绕的区域,并将这些区域里所有的 'O' 用 'X' 填充。

LeetCode —— dfs和bfs_第4张图片

// dfs
class Solution {
    public void solve(char[][] board) {
        for(int i = 0; i < board.length; i++){
            if(board[i][0] == 'O'){
                dfs(board, i, 0);
            }
            if(board[i][board[0].length - 1] == 'O'){
                dfs(board, i, board[0].length - 1);
            }
        }

        for(int j = 1; j < board[0].length - 1; j++){
            if(board[0][j] == 'O'){
                dfs(board, 0, j);
            }
            if(board[board.length - 1][j] == 'O'){
                dfs(board, board.length - 1, j);
            }
        }

        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(board[i][j] == 'A'){
                    board[i][j] = 'O';
                }else if(board[i][j] == 'O'){
                    board[i][j] = 'X';
                }
            }
        }
    }

    public void dfs(char[][] board, int i, int j){
        if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] == 'X' 
             || board[i][j] == 'A'){
            return;
        }
        board[i][j] = 'A';
        dfs(board, i - 1, j);
        dfs(board, i + 1, j);
        dfs(board, i, j - 1);
        dfs(board, i, j + 1);
    }
}
// bfs
class Solution {
    public void solve(char[][] board) {
        for(int i = 0; i < board.length; i++){
            if(board[i][0] == 'O'){
                bfs(board, i, 0);
            }
            if(board[i][board[0].length - 1] == 'O'){
                bfs(board, i, board[0].length - 1);
            }
        }

        for(int j = 1; j < board[0].length - 1; j++){
            if(board[0][j] == 'O'){
                bfs(board, 0, j);
            }
            if(board[board.length - 1][j] == 'O'){
                bfs(board, board.length - 1, j);
            }
        }

        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(board[i][j] == 'A'){
                    board[i][j] = 'O';
                }else if(board[i][j] == 'O'){
                    board[i][j] = 'X';
                }
            }
        }
    }

    public void bfs(char[][] board, int i, int j){
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            i = cur[0];
            j = cur[1];
            if(i >= 0 && i < board.length && j >= 0 && j < board[0].length && board[i][j] == 'O'){
                board[i][j] = 'A';
                queue.offer(new int[]{i - 1, j});
                queue.offer(new int[]{i + 1, j});
                queue.offer(new int[]{i, j - 1});
                queue.offer(new int[]{i, j + 1});
            }
        }
    }
}

 417. 太平洋大西洋水流问题

有一个 m × n 的矩形岛屿,与 太平洋 和 大西洋 相邻。 “太平洋” 处于大陆的左边界和上边界,而 “大西洋” 处于大陆的右边界和下边界。这个岛被分割成一个由若干方形单元格组成的网格。给定一个 m x n 的整数矩阵 heights , heights[r][c] 表示坐标 (r, c) 上单元格 高于海平面的高度 。岛上雨水较多,如果相邻单元格的高度 小于或等于 当前单元格的高度,雨水可以直接向北、南、东、西流向相邻单元格。水可以从海洋附近的任何单元格流入海洋。返回网格标 result 的 2D 列表 ,其中 result[i] = [ri, ci] 表示雨水从单元格 (ri, ci) 流动 既可流向太平洋也可流向大西洋 。

LeetCode —— dfs和bfs_第5张图片

// dfs
class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    public List> pacificAtlantic(int[][] heights) {
        int m = heights.length;
        int n = heights[0].length;
        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];

        for(int i = 0; i < m; i++){
            dfs(heights, pacific, i, 0);
            dfs(heights, atlantic, i, n - 1);
        }
        for(int j = 0; j < n; j++){
            dfs(heights, pacific, 0, j);
            dfs(heights, atlantic, m - 1, j);
        }

        List> list = new ArrayList<>();
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(pacific[i][j] && atlantic[i][j]){
                    list.add(new ArrayList<>(Arrays.asList(i, j)));
                }
            }
        }
        return list;
    }

    public void dfs(int[][] heights, boolean[][] ocean, int i, int j){
        if(ocean[i][j]){
            return;
        }

        ocean[i][j] = true;
        for(int[] pos : positions){
            int newRow = i + pos[0];
            int newColumn = j + pos[1];
            if(newRow < 0 || newRow >= heights.length || newColumn < 0 || newColumn >= heights[0].length || heights[i][j] > heights[newRow][newColumn]){
                continue;
            }
            dfs(heights, ocean, newRow, newColumn);
        }
    }
}
// bfs
class Solution {
    int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    public List> pacificAtlantic(int[][] heights) {
        int m = heights.length;
        int n = heights[0].length;
        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];

        for(int i = 0; i < m; i++){
            bfs(heights, pacific, i, 0);
            bfs(heights, atlantic, i, n - 1);
        }
        for(int j = 0; j < n; j++){
            bfs(heights, pacific, 0, j);
            bfs(heights, atlantic, m - 1, j);
        }

        List> list = new ArrayList<>();
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(pacific[i][j] && atlantic[i][j]){
                    list.add(new ArrayList<>(Arrays.asList(i, j)));
                }
            }
        }
        return list;
    }

    public void bfs(int[][] heights, boolean[][] ocean, int i, int j){
        ocean[i][j] = true;
        Queue queue = new LinkedList<>();
        queue.offer(new int[]{i, j});
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            for(int[] pos : positions){
                int newRow = cur[0] + pos[0];
                int newColumn = cur[1] + pos[1];
                if(newRow < 0 || newRow >= heights.length || newColumn < 0 || newColumn >= heights[0].length || heights[cur[0]][cur[1]] > heights[newRow][newColumn] || ocean[newRow][newColumn]){
                    continue;
                }
                ocean[newRow][newColumn] = true;
                queue.offer(new int[]{newRow, newColumn});
            }
        }
    }
}

841. 钥匙和房间

有 n 个房间,房间按从 0 到 n - 1 编号。最初,除 0 号房间外的其余所有房间都被锁住。你的目标是进入所有的房间。然而,你不能在没有获得钥匙的时候进入锁住的房间。当你进入一个房间,你可能会在里面找到一套不同的钥匙,每把钥匙上都有对应的房间号,即表示钥匙可以打开的房间。你可以拿上所有钥匙去解锁其他房间。给你一个数组 rooms 其中 rooms[i] 是你进入 i 号房间可以获得的钥匙集合。如果能进入 所有 房间返回 true,否则返回 false

示例1:输入:rooms = [[1],[2],[3],[]]         输出:true

示例2:输入:rooms = [[1,3],[3,0,1],[2],[0]]         输出:false

// dfs
class Solution {
    public boolean canVisitAllRooms(List> rooms) {
        boolean[] visited = new boolean[rooms.size()];
        dfs(rooms, visited, 0);
        for(int i = 0; i < visited.length; i++){
            if(!visited[i]){
                return false;
            }
        }
        return true;
    }

    public void dfs(List> rooms, boolean[] visited, int key){
        if(visited[key]){
            return;
        }

        visited[key] = true;
        List curRoom = rooms.get(key);
        for(int i = 0; i < curRoom.size(); i++){
            dfs(rooms, visited, curRoom.get(i));
        }
    }
}
// bfs
class Solution {
    public boolean canVisitAllRooms(List> rooms) {
        boolean[] visited = new boolean[rooms.size()];
        bfs(rooms, visited);
        for(int i = 0; i < visited.length; i++){
            if(!visited[i]){
                return false;
            }
        }
        return true;
    }

    public void bfs(List> rooms, boolean[] visited){
        Queue queue = new LinkedList<>();
        queue.offer(0);
        while(!queue.isEmpty()){
            int curKey = queue.poll();
            visited[curKey] = true;
            List curRoom = rooms.get(curKey);
            for(int i = 0; i < curRoom.size(); i++){
                if(visited[curRoom.get(i)]){
                    continue;
                }
                queue.offer(curRoom.get(i));
            }
        }
    }
}

127. 单词接龙

字典 wordList 中从单词 beginWord 和 endWord 的 转换序列 是一个按下述规格形成的序列 beginWord -> s1 -> s2 -> ... -> sk:①每一对相邻的单词只差一个字母;②对于 1 <= i <= k 时,每个 si 都在 wordList 中。注意, beginWord 不需要在 wordList 中;③sk == endWord。给你两个单词 beginWord 和 endWord 和一个字典 wordList ,返回 从 beginWord 到 endWord 的 最短转换序列 中的 单词数目 。如果不存在这样的转换序列,返回 0 。

示例:输入:beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]         输出:5;

解释:一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。

class Solution {
    public int ladderLength(String beginWord, String endWord, List wordList) {
        Set wordSet = new HashSet<>(wordList);  // 转换为HashSet, 加快速度
        if(!wordSet.contains(endWord)){
            return 0;
        }

        // bfs  广搜只要搜到了终点,那么一定是最短的路径
        Queue queue = new LinkedList<>();
        queue.offer(beginWord);
        Map map = new HashMap<>();  // 存储单词对应的路径长度
        map.put(beginWord, 1);

        while(!queue.isEmpty()){
            String word = queue.poll();
            int pathLen = map.get(word);
            for(int i = 0; i < word.length(); i++){
                char[] chars = word.toCharArray();
                for(char j = 'a'; j <= 'z'; j++){
                    chars[i] = j;
                    String newWord = String.valueOf(chars);
                    if(newWord.equals(endWord)){
                        return pathLen + 1;
                    }
                    if(wordSet.contains(newWord) && !map.containsKey(newWord)){
                        queue.offer(newWord);
                        map.put(newWord, pathLen + 1);
                    }
                }
            }
        }
        return 0;
    }
}

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