一刷1020

Author: diguage

README.adoc

@@ -7165,14 +7165,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/1019-next-greater-node-in-linked-list.adoc[题解]
 //|Medium
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/number-of-enclaves/[1020. Number of Enclaves^]
-//|{source_base_url}/_1020_NumberOfEnclaves.java[Java]
-//|{doc_base_url}/1020-number-of-enclaves.adoc[题解]
-//|Medium
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/number-of-enclaves/[1020. Number of Enclaves^]
+|{source_base_url}/_1020_NumberOfEnclaves.java[Java]
+|{doc_base_url}/1020-number-of-enclaves.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/remove-outermost-parentheses/[1021. Remove Outermost Parentheses^]
 //|{source_base_url}/_1021_RemoveOutermostParentheses.java[Java]

docs/1020-number-of-enclaves.adoc

@@ -1,53 +1,69 @@
 [#1020-number-of-enclaves]
-= 1020. Number of Enclaves
+= 1020. 飞地的数量
 
-{leetcode}/problems/number-of-enclaves/[LeetCode - Number of Enclaves^]
+https://leetcode.cn/problems/number-of-enclaves/[LeetCode - 1020. 飞地的数量 ^]
 
-Given a 2D array `A`, each cell is 0 (representing sea) or 1 (representing land)
+给你一个大小为 `m x n` 的二进制矩阵 `grid` ，其中 `0` 表示一个海洋单元格、`1` 表示一个陆地单元格。
 
-A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.
+一次 *移动* 是指从一个陆地单元格走到另一个相邻（*上、下、左、右*）的陆地单元格或跨过 `grid` 的边界。
 
-Return the number of land squares in the grid for which we *cannot* walk off the boundary of the grid in any number of moves.
+返回网格中 *无法* 在任意次数的移动中离开网格边界的陆地单元格的数量。
 
-
-*Example 1:*
+*示例 1：*
 
 image::images/1020-01.jpg[{image_attr}]
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [[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 isn't enclosed because its on the boundary.
-----
+....
+输入：grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
+输出：3
+解释：有三个 1 被 0 包围。一个 1 没有被包围，因为它在边界上。
+....
 
-*Example 2:*
+*示例 2：*
 
 image::images/1020-02.jpg[{image_attr}]
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [[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.
-
-----
+....
+输入：grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
+输出：0
+解释：所有 1 都在边界上或可以到达边界。
+....
 
+*提示：*
 
-*Note:*
+* `m == grid.length`
+* `n == grid[i].length`
+* `+1 <= m, n <= 500+`
+* `grid[i][j]` 的值为 `0` 或 `1`
 
-. `1 <= A.length <= 500`
-. `1 <= A[i].length <= 500`
-. `0 <= A[i][j] <= 1`
-. All rows have the same size.
 
 == 思路分析
 
+从四条边出发，深度优先遍历，把能出逃的单元格都标记成负数，然后再遍历单元格，统计陆地数量即可。
+
 [[src-1020]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_1020_NumberOfEnclaves.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_1020_NumberOfEnclaves_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
+
 

docs/index.adoc

@@ -2123,7 +2123,7 @@ include::1011-capacity-to-ship-packages-within-d-days.adoc[leveloffset=+1]
 
 // include::1019-next-greater-node-in-linked-list.adoc[leveloffset=+1]
 
-// include::1020-number-of-enclaves.adoc[leveloffset=+1]
+include::1020-number-of-enclaves.adoc[leveloffset=+1]
 
 // include::1021-remove-outermost-parentheses.adoc[leveloffset=+1]
 

logbook/202503.adoc

@@ -823,6 +823,11 @@ endif::[]
 |{doc_base_url}/1033-moving-stones-until-consecutive.adoc[题解]
 |✅ 分类讨论
 
+|{counter:codes2503}
+|{leetcode_base_url}/number-of-enclaves/[1020. 飞地的数量^]
+|{doc_base_url}/1020-number-of-enclaves.adoc[题解]
+|✅ 深度优先遍历。有机会尝试一下广度优先遍历。
+
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_1020_NumberOfEnclaves.java

@@ -4,10 +4,69 @@ public class _1020_NumberOfEnclaves {
   // tag::answer[]
   /**
    * @author D瓜哥 · https://www.diguage.com
-   * @since 2024-07-12 14:37:26
+   * @since 2025-05-25 22:52:15
    */
   public int numEnclaves(int[][] grid) {
-    return 0;
+    int row = grid.length;
+    int col = grid[0].length;
+    for (int i = 0; i < row; i++) {
+      if (grid[i][0] == 1) {
+        dfs(grid, i, 0);
+      }
+      if (grid[i][col - 1] == 1) {
+        dfs(grid, i, col - 1);
+      }
+    }
+    for (int i = 0; i < col; i++) {
+      if (grid[0][i] == 1) {
+        dfs(grid, 0, i);
+      }
+      if (grid[row - 1][i] == 1) {
+        dfs(grid, row - 1, i);
+      }
+    }
+    int result = 0;
+    for (int i = 1; i < row - 1; i++) {
+      for (int j = 1; j < grid[i].length - 1; j++) {
+        if (grid[i][j] == 1) {
+          result++;
+        }
+      }
+    }
+    return result;
+  }
+
+  private void dfs(int[][] grid, int row, int column) {
+    if (row < 0 || row >= grid.length
+      || column < 0 || column >= grid[row].length) {
+      return;
+    }
+    if (grid[row][column] != 1) {
+      return;
+    }
+    grid[row][column] = -1;
+    // 上
+    dfs(grid, row - 1, column);
+    // 下
+    dfs(grid, row + 1, column);
+    // 左
+    dfs(grid, row, column - 1);
+    // 右
+    dfs(grid, row, column + 1);
   }
   // end::answer[]
+  public static void main(String[] args) {
+    new _1020_NumberOfEnclaves().numEnclaves(new int[][]{
+      {0, 0, 0, 1, 1, 1, 0, 1, 0, 0},
+      {1, 1, 0, 0, 0, 1, 0, 1, 1, 1},
+      {0, 0, 0, 1, 1, 1, 0, 1, 0, 0},
+      {0, 1, 1, 0, 0, 0, 1, 0, 1, 0},
+      {0, 1, 1, 1, 1, 1, 0, 0, 1, 0},
+      {0, 0, 1, 0, 1, 1, 1, 1, 0, 1},
+      {0, 1, 1, 0, 0, 0, 1, 1, 1, 1},
+      {0, 0, 1, 0, 0, 1, 0, 1, 0, 1},
+      {1, 0, 1, 0, 1, 1, 0, 0, 0, 0},
+      {0, 0, 0, 0, 1, 1, 0, 0, 0, 1}
+    });
+  }
 }