797.所有可能的路径：
. - 力扣（LeetCode）
class Solution {List<List<Integer>> ans=new LinkedList<>();List<Integer> list=new LinkedList<>();public List<List<Integer>> allPathsSourceTarget(int[][] graph) {list.add(0);backTracking(graph,0);return ans;}private void backTracking(int[][] graph,int x){if (x==graph.length-1){ans.add(new LinkedList<>(list));}for (int i = 0; i < graph[x].length; i++) {list.add(graph[x][i]);backTracking(graph,graph[x][i]);list.remove(list.size()-1);}}
}岛屿数量：
. - 力扣（LeetCode）
深搜：
class Solution {int ans=0;int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};public int numIslands(char[][] grid) {for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j]=='1'){ans++;grid[i][j]='0';backTracking(grid,i,j);}}}return ans;}private void backTracking(char[][] grid,int x,int y){for (int i = 0; i < 4; i++) {int newx=x+dx[i];int newy=y+dy[i];if (newx<0||newy<0||newx>=grid.length||newy>=grid[0].length||grid[newx][newy]=='0')continue;grid[newx][newy]='0';backTracking(grid,newx,newy);}}
}广搜：
class Solution {int ans=0;int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};public int numIslands(char[][] grid) {for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j]=='1'){ans++;bfs(grid,new boolean[grid.length][grid[0].length],i,j);}}}return ans;}private void bfs(char[][] grid,boolean[][] vis,int x,int y) {Deque<Pair> pairDeque=new LinkedList<>();pairDeque.push(new Pair(x,y));while (!pairDeque.isEmpty()){Pair pair = pairDeque.pop();int x1 = pair.x;int y1 = pair.y;for (int i = 0; i < 4; i++) {int newx=x1+dx[i];int newy=y1+dy[i];if (newx<0||newy<0||newx>=grid.length||newy>=grid[0].length||grid[newx][newy]=='0'){continue;}grid[newx][newy]='0';pairDeque.push(new Pair(newx,newy));}}}class Pair{int x;int y;public Pair(int x, int y) {this.x = x;this.y = y;}}
}岛屿的最大面积：
. - 力扣（LeetCode）
int max=0;int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};int count=1;public int maxAreaOfIsland(int[][] grid) {for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j]==1){count=0;dfs(grid,i,j);max=Math.max(max,count);}}}return max;}private void dfs(int[][] grid,int x,int y){count++;grid[x][y]=0;for (int i = 0; i < 4; i++) {int newx=x+dx[i];int newy=y+dy[i];if (newx<0||newy<0||newx>=grid.length||newy>=grid[0].length||grid[newx][newy]==0){continue;}grid[newx][newy]=0;dfs(grid,newx,newy);}}public static void main(String[] args) {LeetCode695dfs leetCode695=new LeetCode695dfs();int max = leetCode695.maxAreaOfIsland(new int[][]{{1}});System.out.println(max);}class Solution {int max=0;int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};int count=1;public int maxAreaOfIsland(int[][] grid) {for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j]==1){count=0;bfs(grid,i,j);max=Math.max(count,max);}}}return max;}private void bfs(int[][] grid,int x,int y){Deque<pair> pairDeque=new LinkedList<>();pairDeque.push(new pair(x,y));grid[x][y]=0;while (!pairDeque.isEmpty()){pair pop = pairDeque.pop();count++;int x1 = pop.x;int y1 = pop.y;for (int i = 0; i < 4; i++) {int newx=x1+dx[i];int newy=y1+dy[i];if (newx<0||newy<0||newx>=grid.length||newy>=grid[0].length||grid[newx][newy]==0){continue;}grid[newx][newy]=0;pairDeque.push(new pair(newx,newy));}}}class pair{int x;int y;public pair(int x, int y) {this.x = x;this.y = y;}}
}1020：飞地的数量：
. - 力扣（LeetCode）
class Solution {int ans=0;int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};boolean flag=true;public int numEnclaves(int[][] grid) {for (int i = 0; i < grid[0].length; i++) {if (grid[0][i]==1){grid[0][i]=0;dfs(grid,0,i);}if (grid[grid.length-1][i]==1){grid[grid.length-1][i]=0;dfs(grid,grid.length-1,i);}}for (int i = 0; i < grid.length; i++) {if (grid[i][0]==1){grid[i][0]=0;dfs(grid,i,0);}if (grid[i][grid[0].length-1]==1){grid[i][grid[0].length-1]=0;dfs(grid,i,grid[0].length-1);}}for (int i = 1; i < grid.length-1; i++) {for (int j = 1; j < grid[0].length-1; j++) {if (grid[i][j]==1) ans++;}}return ans;}private void dfs(int[][] grid,int x,int y){for (int i = 0; i < 4; i++) {int newx=x+dx[i];int newy=y+dy[i];if (newx<0||newy<0||newx>=grid.length||newy>=grid[0].length||grid[newx][newy]==0){continue;}grid[newx][newy]=0;dfs(grid,newx,newy);}}
}这题广搜更胜一筹：
public class LeetCode1020bfs {int ans=0;int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};public int numEnclaves(int[][] grid) {for (int i = 0; i < grid[0].length; i++) {if (grid[0][i]==1){grid[0][i]=0;bfs(grid,0,i);}if (grid[grid.length-1][i]==1){grid[grid.length-1][i]=0;bfs(grid,grid.length-1,i);}}for (int i = 0; i < grid.length; i++) {if (grid[i][0]==1){grid[i][0]=0;bfs(grid,i,0);}if (grid[i][grid[0].length-1]==1){grid[i][grid[0].length-1]=0;bfs(grid,i,grid[0].length-1);}}for (int i = 1; i < grid.length-1; i++) {for (int j = 1; j < grid[0].length-1; j++) {if (grid[i][j]==1) ans++;}}return ans;}private void bfs(int[][] grid,int x,int y){Deque<pair> deque=new LinkedList<>();deque.push(new pair(x,y));grid[x][y]=0;while (!deque.isEmpty()){pair pop = deque.pop();int x1 = pop.x;int y1 = pop.y;for (int i = 0; i < 4; i++) {int newx=x1+dx[i];int newy=y1+dy[i];if (newx<0||newy<0||newx>=grid.length||newy>=grid[0].length||grid[newx][newy]==0){continue;}grid[newx][newy]=0;deque.push(new pair(newx,newy));}}}class pair{int x;int y;public pair(int x, int y) {this.x = x;this.y = y;}}
}
被围绕的区域：
. - 力扣（LeetCode）
class Solution {int[] dx={0,1,-1,0};int[] dy={1,0,0,-1};public void solve(char[][] board) {int[][] flag=new int[board.length][board[0].length];for (int i = 0; i < board.length; i++) {if (board[i][0]=='O'){dfs(board,i,0,flag);}if (board[i][board[0].length-1]=='O'){dfs(board,i,board[0].length-1,flag);}}for (int i = 0; i < board[0].length; i++) {if (board[0][i]=='O'){dfs(board,0,i,flag);}if (board[board.length-1][i]=='O'){dfs(board,board.length-1,i,flag);}}for (int i = 1; i <board.length-1 ; i++) {for (int j = 1; j < board[0].length-1; j++) {if (flag[i][j]==0&&board[i][j]=='O'){board[i][j]='X';}}}}private void dfs(char[][] board, int x, int y, int[][] flag) {flag[x][y]=1;for (int i = 0; i < 4; i++) {int newx=x+dx[i];int newy=y+dy[i];if (newx<0||newy<0||newx>=board.length||newy>=board[0].length||board[newx][newy]=='X'||flag[newx][newy]==1){continue;}dfs(board,newx,newy,flag);}}
}
)
