solve : #258 (Number of Islands) with Java

Author: kdh-92

number-of-islands/kdh-92.java

@@ -0,0 +1,45 @@
+class Solution {
+    public int numIslands(char[][] grid) {
+        // bfs
+        // 시간복잡도 : O(r * c), 공간복잡도 O(r * c)
+        // 풀이
+        // bfs 풀이에 방문을 Set<String>으로 설정하여 체크, directions(상하좌우)를 통해 탐색
+        // 응용 가능 : 대각선 추가하여 연결된 섬 체크
+        int islands = 0;
+        int rows = grid.length;
+        int cols = grid[0].length;
+        Set<String> visited = new HashSet<>();
+
+        int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
+
+        for (int r = 0; r < rows; r++) {
+            for (int c = 0; c < cols; c++) {
+                if (grid[r][c] == '1' && !visited.contains(r + "," + c)) {
+                    islands++;
+                    bfs(grid, r, c, visited, directions, rows, cols);
+                }
+            }
+        }
+
+        return islands;
+    }
+
+    private void bfs(char[][] grid, int r, int c, Set<String> visited, int[][] directions, int rows, int cols) {
+        Queue<int[]> q = new LinkedList<>();
+        visited.add(r + "," + c);
+        q.add(new int[]{r, c});
+
+        while (!q.isEmpty()) {
+            int[] point = q.poll();
+            int row = point[0], col = point[1];
+
+            for (int[] direction : directions) {
+                int nr = row + direction[0], nc = col + direction[1];
+                if (nr >= 0 && nr < rows && nc >= 0 && nc < cols && grid[nr][nc] == '1' && !visited.contains(nr + "," + nc)) {
+                    q.add(new int[] {nr, nc});
+                    visited.add(nr + "," + nc);
+                }
+            }
+        }
+    }
+}