day-101 飞地的数量

思路
DFS深度优先搜索

解题过程
从矩阵的四周边界中的陆地进行深度优先搜索，将访问过的陆地进行标记，所有边界位置都进行尝试后，还没被访问过的陆地数即为飞地的数量

Code

class Solution {
    int dir[][] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
    static int vis[][];
    int row;
    int col;

    public int numEnclaves(int[][] grid) {
        int ans = 0;
        row = grid.length;
        col = grid[0].length;
        vis = new int[row][col];
        for (int i = 0; i < row; i++) {
            if (grid[i][0] == 1 && vis[i][0] == 0)
                dfs(i, 0, grid);
            if (grid[i][col - 1] == 1 && vis[i][col - 1] == 0)
                dfs(i, col - 1, grid);
        }

        for (int i = 0; i < col; i++) {
            if (grid[0][i] == 1 && vis[0][i] == 0)
                dfs(0, i, grid);
            if (grid[row - 1][i] == 1 && vis[row - 1][i] == 0)
                dfs(row - 1, i, grid);
        }

        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == 1 && vis[i][j] == 0)
                    ans++;
            }
        }
        return ans;
    }

    public void dfs(int x, int y, int[][] grid) {
        vis[x][y] = 1;
        for (int i = 0; i < 4; i++) {
            int newx = x + dir[i][0];
            int newy = y + dir[i][1];
            if (0 <= newx && newx < row && 0 <= newy && newy < col && vis[newx][newy] == 0 && grid[newx][newy] == 1) {
                dfs(newx, newy, grid);
            }
        }
    }
}

作者：菜卷
链接：https://leetcode.cn/problems/number-of-enclaves/solutions/3032932/fei-di-de-shu-liang-by-ashi-jian-chong-d-nuvr/
来源：力扣（LeetCode）
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