首頁 解題筆記 Leetcode Medium

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# 題目敘述

Given a 2D grid consists of 0s (land) and 1s (water). An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s .

Return the number of closed islands.

# Example 1:

Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation:
Islands in gray are closed because they are completely surrounded by water (group of 1s).

# Example 2:

Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
Output: 1

# Example 3:

Input: grid = [[1,1,1,1,1,1,1],[1,0,0,0,0,0,1],[1,0,1,1,1,0,1],[1,0,1,0,1,0,1],
[1,0,1,1,1,0,1],[1,0,0,0,0,0,1],[1,1,1,1,1,1,1]]
Output: 2

# 解題思路

利用 dfs(Depth-First Search) 去檢查如果該陣列為 0 ，他前後左右是否會碰到 1 ，如果碰到邊界表示封閉，如果碰到 0 再繼續找。

# Solution

class Solution {
    public int closedIsland(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        boolean[][] visit = new boolean[m][n];
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0 && !visit[i][j]) {
                    if(dfs(i, j, m, n, grid, visit)) ans++;
                }
            }
        }
        return ans;
    }
    public boolean dfs(int x, int y, int m, int n, int[][] grid, boolean[][] visit) {
        if (x < 0 || x >= m || y < 0 || y >= n) {
            return false;
        }
        if (grid[x][y] == 1 || visit[x][y]) {
            return true;
        }
        visit[x][y] = true;
        boolean isClosed = true;
        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];
            isClosed &= dfs(r, c, m, n, grid, visit);
        }
        return isClosed;
    }
}