Update LeetCode Post " LeetCode : 200. Number of Islands "

_posts/2021-12-05-LeetCode-Number-of-Islands.md

@@ -98,7 +98,95 @@ class Solution {
 
 * 단순 구현 + BFS 풀이
 
+---
+
+### [2] Code (24. 03. 01)
+
+*Retry -> 풀긴했는데 코드가 더럽다.*
+
+``` java
+// Runtime: 9 ms
+// Memory Usage: 54.1 MB
+// Ref : https://leetcode.com/submissions/detail/1190639057
+class Solution {
+    public int numIslands(char[][] _grid) {
+        int[] dx = { 1, -1, 0, 0 };
+        int[] dy = { 0, 0, 1, -1 };
+
+        int n = _grid.length;
+        int m = _grid[0].length;
+
+        Queue<int[]> q = new LinkedList<>();
+        int[][] grid = new int[n][m];
+        int[][] visit = new int[n][m];
+
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                grid[i][j] = _grid[i][j] - '0';
+                if (grid[i][j] == 1) {
+                    q.add(new int[] { i, j });
+                }
+            }
+        }
+
+        int cnt = 2;
+
+        while (!q.isEmpty()) {
+            int[] _node = q.poll();
+            int _x = _node[0];
+            int _y = _node[1];
+
+            if (grid[_x][_y] != 1) {
+                continue;
+            }
+            grid[_x][_y] = cnt;
+            Queue<int[]> q2 = new LinkedList<>();
+            q2.add(new int[] { _x, _y });
+
+            while (!q2.isEmpty()) {
+                int[] node = q2.poll();
+                int x = node[0];
+                int y = node[1];
+
+                for (int i = 0; i < 4; i++) {
+                    int nx = x + dx[i];
+                    int ny = y + dy[i];
+
+                    if (!isLand(grid, nx, ny)) {
+                        continue;
+                    }
+                    grid[nx][ny] = cnt;
+                    q2.add(new int[] { nx, ny });
+                }
+            }
+            cnt++;
+        }
+        return cnt - 2;
+    }
+
+    private boolean isLand(int[][] map, int x, int y) {
+        int n = map.length;
+        int m = map[0].length;
+        if (x < 0 || x >= n || y < 0 || y >= m) {
+            return false;
+        }
+
+        if (map[x][y] == 0 || map[x][y] != 1) {
+            return false;
+        }
+
+        return true;
+    }
+}
+```
+
+* 풀긴했는데 코드가 깔끔하지 못해서 마음에 안든다.
+
+  [1]에 풀었던 코드를 보니 훨씬 깔끔하다.
+
+  ~~코드를 깔끔하게 짜는 설계 능력이 퇴보했다.~~
 
+* 위안을 삼자면 그나마 제출 1번에 바로 맞췄다.
 
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