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文章目录

• 前言
• 一、力扣200. 岛屿数量
• 二、力扣695. 岛屿的最大面积
• 三、力扣1020. 飞地的数量
• 四、力扣130. 被围绕的区域

  ---

  前言

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  依然是从地图周边出发，将周边空格相邻的'O'都做上标记，然后在遍历一遍地图，遇到 'O' 且没做过标记的，那么都是地图中间的'O'，全部改成'X'就行。

  一、力扣200. 岛屿数量

  class Solution {
      int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
      boolean[][] flag;
      public int numIslands(char[][] grid) {
          int m = grid.length, n = grid[0].length;
          int res = 0;
          flag = new boolean[m][n];
          for(int i = 0; i < m; i ++){
              for(int j = 0; j < n; j ++){
                  if(!flag[i][j] && grid[i][j] == '1'){
                      res ++;
                      bfs(grid, i, j);
                  }
              }
          }
          return res;
      }
      public void bfs(char[][] grid, int x, int y){
          Deque deq = new LinkedList<>();
          deq.offerLast(new int[]{x,y});
          while(!deq.isEmpty()){
              int size = deq.size();
              for(int i = 0; i < size; i ++){
                  int[] cur = deq.pollFirst();
                  for(int j = 0; j < 4; j ++){
                      int curX = cur[0] + arr[j][0];
                      int curY = cur[1] + arr[j][1];
                      if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
                          continue;
                      }
                      if(!flag[curX][curY] && grid[curX][curY] == '1'){
                          flag[curX][curY] = true;
                          deq.offerLast(new int[]{curX,curY});
                      }
                  }
              }
          }
      }
  }
  

  二、力扣695. 岛屿的最大面积

  class Solution {
      int res = 0;
      int count = 0;
      int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
      boolean[][] flag;
      public int maxAreaOfIsland(int[][] grid) {
          int m = grid.length, n = grid[0].length;
          flag = new boolean[m][n];
          for(int i = 0; i < m; i ++){
              for(int j = 0; j < n; j ++){
                  if(!flag[i][j] && grid[i][j] == 1){
                      flag[i][j] = true;
                      count = 1;
                      dfs(grid,i,j);
                  }
              }
          }
          return res;
      }
      public void dfs(int[][] grid, int x, int y){
          res = Math.max(count,res);
          for(int i = 0; i < 4; i ++){
              int curX = x + arr[i][0];
              int curY = y + arr[i][1];
              if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
                  continue;
              }
              if(!flag[curX][curY] && grid[curX][curY] == 1){
                  flag[curX][curY] = true;
                  count ++;
                  dfs(grid,curX,curY);
              }
          }
      }
  }
  

  三、力扣1020. 飞地的数量

  class Solution {
      int res = 0;
      int count = 0;
      int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
      boolean[][] flag;
      boolean f;
      public int numEnclaves(int[][] grid) {
          int m = grid.length, n = grid[0].length;
          flag = new boolean[m][n];
          for(int i = 0; i < m; i ++){
              for(int j = 0; j < n; j ++){
                  if(!flag[i][j] && grid[i][j] == 1){
                      flag[i][j] = true;
                      count = 1;
                      f = false;
                      dfs(grid,i,j);
                      if(!f){
                          res += count;
                      }
                  }
              }
          }
          return res;
      }
      public void dfs(int[][] grid, int x, int y){
          for(int i = 0; i < 4; i ++){
              int curX = x + arr[i][0];
              int curY = y + arr[i][1];
              if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
                  f = true;
                  continue;
              }
              if(!flag[curX][curY] && grid[curX][curY] == 1){
                  count ++;
                  flag[curX][curY] = true;
                  dfs(grid,curX,curY);
              }
          }
      }
  }
  

  四、力扣130. 被围绕的区域

  class Solution {
      int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
      boolean[][] flag;
      public void solve(char[][] board) {
          int m = board.length, n = board[0].length;
          flag = new boolean[m][n];
          for(int i = 0; i < m; i ++){
              if(!flag[i][0] && board[i][0] == 'O'){
                  flag[i][0] = true;
                  dfs(board, i, 0);
              }
              if(!flag[i][n-1] && board[i][n-1] == 'O'){
                  flag[i][n-1] = true;
                  dfs(board,i,n-1);
              }
          }
          for(int i = 0; i < n; i ++){
              if(!flag[0][i] && board[0][i] == 'O'){
                  flag[0][i] = true;
                  dfs(board,0,i);
              }
              if(!flag[m-1][i] && board[m-1][i] == 'O'){
                  flag[m-1][i] = true;
                  dfs(board,m-1,i);
              }
          }
          for(int i = 0; i < m; i ++){
              for(int j = 0; j < n; j ++){
                  if(!flag[i][j] && board[i][j] == 'O'){
                      board[i][j] = 'X';
                  }
              }
          }
      }
      public void dfs(char[][] board, int x, int y){
          for(int i = 0; i < 4; i++){
              int curX = x + arr[i][0];
              int curY = y + arr[i][1];
              if(curX < 0 || curX >= board.length || curY < 0 || curY >= board[0].length){
                  continue;
              }
              if(!flag[curX][curY] && board[curX][curY] == 'O'){
                  flag[curX][curY] = true;
                  dfs(board,curX,curY);
              }
          }
      }
  }