add number of islands solution

Author: Tessa1217

number-of-islands/Tessa1217.java

@@ -0,0 +1,67 @@
+import java.util.LinkedList;
+import java.util.Queue;
+
+/**
+ * m x n 2D 행렬 grid가 주어질 때 섬의 수를 찾아서 반환하세요.
+ * 섬(island)는 물(0)로 둘러싸여 있고 가로 또는 세로로 인접한 섬들과 연결되어 형성되어 있다.
+ */
+class Solution {
+
+    int[] dx = {-1, 1, 0, 0};
+    int[] dy = {0, 0, 1, -1};
+
+    public int numIslands(char[][] grid) {
+        int cnt = 0;
+        for (int i = 0; i < grid.length; i++) {
+            for (int j = 0; j < grid[i].length; j++) {
+                //  land라면
+                if (grid[i][j] == '1') {
+
+                    // bfs로 섬 탐색
+                    bfs(i, j, grid);
+
+                    // dfs로 섬 탐색
+                    dfs(i, j, grid);
+
+                    cnt++;
+                }
+            }
+        }
+        return cnt;
+    }
+
+    private void dfs(int x, int y, char[][] grid) {
+        if (grid[x][y] == '1') {
+            grid[x][y] = '0';
+        }
+        for (int i = 0; i < 4; i++) {
+            int nx = x + dx[i];
+            int ny = y + dy[i];
+            if (nx >= 0 && nx < grid.length && ny >= 0 && ny < grid[0].length && grid[nx][ny] == '1') {
+                dfs(nx, ny, grid);
+            }
+        }
+    }
+
+    private void bfs(int startX, int startY, char[][] grid) {
+
+        Queue<int[]> queue = new LinkedList<>();
+        queue.offer(new int[]{startX, startY});
+        grid[startX][startY] = '0';
+
+        while (!queue.isEmpty()) {
+            int[] current = queue.poll();
+            for (int i = 0; i < 4; i++) {
+                int newX = current[0] + dx[i];
+                int newY = current[1] + dy[i];
+                if (newX >= 0 && newX < grid.length && newY >= 0 && newY < grid[0].length) {
+                    if (grid[newX][newY] == '1') {
+                        queue.offer(new int[]{newX, newY});
+                        grid[newX][newY] = '0';
+                    }
+                }
+            }
+        }
+    }
+}
+