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1254:统计封闭岛屿的数目

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src/main/java/leetcode/editor/cn/NumberOfClosedIslands.java Normal file
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//有一个二维矩阵 grid ，每个位置要么是陆地（记号为 0 ）要么是水域（记号为 1 ）。
//
// 我们从一块陆地出发，每次可以往上下左右 4 个方向相邻区域走，能走到的所有陆地区域，我们将其称为一座「岛屿」。
//
// 如果一座岛屿 完全 由水域包围，即陆地边缘上下左右所有相邻区域都是水域，那么我们将其称为 「封闭岛屿」。
//
// 请返回封闭岛屿的数目。
//
//
//
// 示例 1：
//
//
//
// 输入：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]]
//输出：2
//解释：
//灰色区域的岛屿是封闭岛屿，因为这座岛屿完全被水域包围（即被 1 区域包围）。
//
// 示例 2：
//
//
//
// 输入：grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
//输出：1
//
//
// 示例 3：
//
// 输入：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]]
//输出：2
//
//
//
//
// 提示：
//
//
// 1 <= grid.length, grid[0].length <= 100
// 0 <= grid[i][j] <=1
//
// Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵
// 👍 83 👎 0
package leetcode.editor.cn;
import com.code.leet.entiy.TwoArray;
//1254:统计封闭岛屿的数目
public class NumberOfClosedIslands {
public static void main(String[] args) {
//测试代码
Solution solution = new NumberOfClosedIslands().new Solution();
TwoArray twoArray = new TwoArray("" +
"[[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]]"
);
System.out.println(solution.closedIsland(twoArray.getArr()));
}
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
boolean[][] use;
int[] dx = new int[]{-1, 1, 0, 0};
int[] dy = new int[]{0, 0, -1, 1};
int count = 0;
boolean flag;
public int closedIsland(int[][] grid) {
use = new boolean[grid.length][grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 0 && !use[i][j]) {
flag = true;
dfs(i, j, grid);
if (flag) {
count++;
}
}
}
}
return count;
}
private void dfs(int x, int y, int[][] grid) {
use[x][y] = true;
for (int i = 0; i < 4; i++) {
x += dx[i];
y += dy[i];
if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
x -= dx[i];
y -= dy[i];
flag = false;
continue;
}
if (grid[x][y] == 0 && !use[x][y]) {
dfs(x, y, grid);
}
x -= dx[i];
y -= dy[i];
}
}
}
//leetcode submit region end(Prohibit modification and deletion)
}

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src/main/java/leetcode/editor/cn/NumberOfClosedIslands.md Normal file
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<p>有一个二维矩阵 <code>grid</code>&nbsp;，每个位置要么是陆地（记号为&nbsp;<code>0</code> ）要么是水域（记号为&nbsp;<code>1</code> ）。</p>
<p>我们从一块陆地出发，每次可以往上下左右&nbsp;4 个方向相邻区域走，能走到的所有陆地区域，我们将其称为一座「<strong>岛屿</strong>」。</p>
<p>如果一座岛屿&nbsp;<strong>完全</strong>&nbsp;由水域包围，即陆地边缘上下左右所有相邻区域都是水域，那么我们将其称为 「<strong>封闭岛屿</strong>」。</p>
<p>请返回封闭岛屿的数目。</p>
<p>&nbsp;</p>
<p><strong>示例 1：</strong></p>
<p><img src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/11/07/sample_3_1610.png"></p>
<pre><strong>输入：</strong>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]]
<strong>输出：</strong>2
<strong>解释：</strong>
灰色区域的岛屿是封闭岛屿，因为这座岛屿完全被水域包围（即被 1 区域包围）。</pre>
<p><strong>示例 2：</strong></p>
<p><img src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/11/07/sample_4_1610.png"></p>
<pre><strong>输入：</strong>grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
<strong>输出：</strong>1
</pre>
<p><strong>示例 3：</strong></p>
<pre><strong>输入：</strong>grid = [[1,1,1,1,1,1,1],
&nbsp; [1,0,0,0,0,0,1],
&nbsp; [1,0,1,1,1,0,1],
&nbsp; [1,0,1,0,1,0,1],
&nbsp; [1,0,1,1,1,0,1],
&nbsp; [1,0,0,0,0,0,1],
[1,1,1,1,1,1,1]]
<strong>输出：</strong>2
</pre>
<p>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 &lt;= grid.length, grid[0].length &lt;= 100</code></li>
<li><code>0 &lt;= grid[i][j] &lt;=1</code></li>
</ul>
<div><div>Related Topics</div><div><li>深度优先搜索</li><li>广度优先搜索</li><li>并查集</li><li>数组</li><li>矩阵</li></div></div>\n<div><li>👍 83</li><li>👎 0</li></div>