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# 題目敘述

You are given an m x n binary matrix grid , where 0 represents a sea cell and 1 represents a land cell.

A move consists of walking from one land cell to another adjacent (4-directionally) land cell or walking off the boundary of the grid .

Return the number of land cells in grid for which we cannot walk off the boundary of the grid in any number of moves.

# Example 1:

Input: grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation: There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.

# Example 2:

Input: grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation: All 1s are either on the boundary or can reach the boundary.

# 解題思路

# Solution

class Solution {
    public int numEnclaves(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if ((i == 0 || j == 0 || i == m - 1 || j == n - 1) && grid[i][j] == 1) {
                    dfs(i, j, m, n, grid);
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    ans++;
                }
            }
        }
        return ans;
    }
    public void dfs(int x, int y, int m, int n, int[][] grid) {
        grid[x][y] = 0;
        int[] dirx = { 0, 1, 0, -1 };
        int[] diry = { -1, 0, 1, 0 };
        for (int i = 0; i < 4; i++) {
            int r = x + dirx[i];
            int c = y + diry[i];
            if (r >= 0 && r < m && c >= 0 && c < n && grid[r][c] == 1) {
                dfs(r, c, m, n, grid);
            }
        }
    }
}




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