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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<Integer>> list = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> 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
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);}
}

// bfs
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<int[]> 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});}}}
}

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 。

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<int[]> 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});}}}
}

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。

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<int[]> 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。

// 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<int[]> 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});}}}
}