原题链接：

1020. 飞地的数量

https://leetcode.cn/problems/number-of-enclaves/description/

完成情况：

解题思路：

private final static int [][] directions = {{-1,0},{1,0},{0,-1},{0,1}};
private int rows,columns; //减少递归函数中，额外的参数调用
private boolean [][] visited;

/**
*

• @param grid
• @return
  */
  … for (int i = 1;i<rows-1;i++){
  for (int j = 1;j<columns-1;j++){
  if (grid[i][j] == 1 && !visited[i][j]){
  enclaves++;
  }
  }
  }
  return enclaves;
  }“”

这段代码定义了一个二维数组 directions，存储了四个方向的偏移量；声明了 rows 和 columns 变量，用来减少递归函数中的额外参数调用；还定义了一个二维布尔数组 visited。

在主要的代码逻辑中，通过两个 for 循环遍历二维数组 grid，当 grid[i][j] 的值为 1 且 visited[i][j] 为 false 时，将 enclaves 的值加一。最终返回 enclaves 的值。

参考代码：

_1020飞地的数量_dfs_定义方向

package leetcode板块;public class _1020飞地的数量_dfs_定义方向 {private final static int [][] directions = {{-1,0},{1,0},{0,-1},{0,1}};private int rows,columns;   //减少递归函数中，额外的参数调用private boolean [][] visited;/**** @param grid* @return*/public int numEnclaves(int[][] grid) {rows = grid.length;columns = grid[0].length;visited = new boolean[rows][columns];//----------先处理四个边界---------------------------for (int i = 0; i < rows; i++){dfs_numEnclaves(grid,i,0);dfs_numEnclaves(grid,i,columns-1);}for (int j = 1;j<columns;j++){dfs_numEnclaves(grid,0,j);dfs_numEnclaves(grid,rows-1,j);}//--------对中间值逐个去dfs，寻找满足条件的飞地enclaves---------int enclaves = 0;for (int i = 1;i<rows - 1;i++){for (int j = 1;j<columns - 1;j++){if (grid[i][j] == 1 && !visited[i][j]){     //如果满足为1，且已经访问过这个节点，那么它就是飞地enclavesenclaves++;}}}return enclaves;}/**** @param grid* @param curRow* @param curCol*/private void dfs_numEnclaves(int[][] grid, int curRow, int curCol) {//边界判断if (curRow < 0 || curRow >= rows || curCol < 0 || curCol >= columns || grid[curRow][curCol] == 0 || visited[curRow][curCol]){return ;}//节点为1visited[curRow][curCol] = true;//遍历它的四个方向for (int [] dir :directions){dfs_numEnclaves(grid,curRow + dir[0],curCol+dir[1]);}}
}

_1020飞地的数量_bfs_定义方向

package leetcode板块;import java.util.ArrayDeque;
import java.util.Deque;public class _1020飞地的数量_bfs_定义方向 {private final static int [][] directions = {{-1,0},{1,0},{0,-1},{0,1}};private int rows,columns;   //减少递归函数中，额外的参数调用private boolean [][] visited;/**** @param grid* @return*/public int numEnclaves(int[][] grid) {rows = grid.length;columns = grid[0].length;visited = new boolean[rows][columns];//bfs就不需要额外调用了，使用队列来解决Deque<int []> myQueue = new ArrayDeque<>();     //用于判断四个方向，所以采用int []//----------先处理四个边界---------------------------for (int i = 0; i < rows; i++){if (grid[i][0] == 1){visited[i][0] = true;myQueue.offer(new int[]{i,0});}if (grid[i][columns - 1] == 1){visited[i][columns - 1] = true;myQueue.offer(new int[]{i,columns-1});}}for (int j = 1;j<columns;j++){if (grid[0][j] == 1){visited[0][j] = true;myQueue.offer(new int[]{0,j});}if (grid[rows-1][j] == 1){visited[rows-1][j] = true;myQueue.offer(new int[]{rows-1,j});}}//-------------对队列中每一个元素进行四个方向的遍历查询----------------------while (!myQueue.isEmpty()){int [] queueCell  = myQueue.poll();int curRow = queueCell[0],curCol = queueCell[1];for (int [] dir : directions){int nextRow = curRow + dir[0],nextCol = curCol + dir[1];if (nextRow >= 0 && nextRow < rows && nextCol >= 0 && nextCol < columns && grid[nextRow][nextCol] == 1 && !visited[nextRow][nextCol]){visited[nextRow][nextCol] = true;myQueue.offer(new int[]{nextRow,nextCol});}}}int enclaves = 0;for (int i = 1;i<rows-1;i++){for (int j = 1;j<columns-1;j++){if (grid[i][j] == 1 && !visited[i][j]){enclaves++;}}}return enclaves;}}

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