From 53a4a17b8c518b6e27166b3b9b60de44fdf9b39a Mon Sep 17 00:00:00 2001
From: Camille <59353274+Camille0512@users.noreply.github.com>
Date: Thu, 26 Sep 2024 23:06:45 +0800
Subject: [PATCH] =?UTF-8?q?Update=200099.=E5=B2=9B=E5=B1=BF=E7=9A=84?=
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 =?UTF-8?q?=20for=20the=20counting=20island=20puzzle=20using=20BFS=20algo?=
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---
 .../kamacoder/0099.岛屿的数量广搜.md   | 49 +++++++++++++++++++
 1 file changed, 49 insertions(+)

diff --git a/problems/kamacoder/0099.岛屿的数量广搜.md b/problems/kamacoder/0099.岛屿的数量广搜.md
index 30475753..9d31c922 100644
--- a/problems/kamacoder/0099.岛屿的数量广搜.md
+++ b/problems/kamacoder/0099.岛屿的数量广搜.md
@@ -499,6 +499,55 @@ main();
 ### Swift
 
 ### Scala
+```scala
+import scala.collection.mutable.Queue
+import util.control.Breaks._
+
+// Dev on LeetCode: https://leetcode.cn/problems/number-of-islands/description/
+object Solution {
+    def numIslands(grid: Array[Array[Char]]): Int = {
+        val row = grid.length
+        val col = grid(0).length
+        val dir = List((-1,0), (0,-1), (1,0), (0,1)) // 四个方向
+        var visited = Array.fill(row)(Array.fill(col)(false))
+        var counter = 0
+        var que = Queue.empty[Tuple2[Int, Int]]
+
+        (0 until row).map{ r =>
+            (0 until col).map{ c =>
+                breakable {
+                    if (!visited(r)(c) && grid(r)(c) == '1') {
+                        que.enqueue((r, c))
+                        visited(r)(c) // 只要加入队列，立刻标记
+                    } else break // 不是岛屿不进入queue，也不记录
+
+                    while (!que.isEmpty) {
+                        val cur = que.head
+                        que.dequeue()
+                        val x = cur(0)
+                        val y = cur(1)
+                        dir.map{ d =>
+                            val nextX = x + d(0)
+                            val nextY = y + d(1)
+                            breakable {
+                                // 越界就跳过
+                                if (nextX < 0 || nextX >= row || nextY < 0 || nextY >= col) break
+                                if (!visited(nextX)(nextY) && grid(nextX)(nextY) == '1') {
+                                    visited(nextX)(nextY) = true // 只要加入队列，立刻标记
+                                    que.enqueue((nextX, nextY))
+                                }
+                            }
+                        }
+                    }
+                    counter = counter + 1 // 找完一个岛屿后记录一下
+                }
+            }
+        }
+        
+        counter
+    }
+}
+```
 
 ### C#