解決跑板問題 + 提供java解法(和卡哥邏輯一致)

跑版: 原本的code box 結束的```不見了，所以看不到結尾圖片。

新解法和原本JAVA解法的差異: 原版的應該是自己寫的DFS，邏輯大致一致，但還是有差異，故提供用卡哥C++ code改的版本

problems/0200.岛屿数量.深搜版.md

@@ -176,6 +176,48 @@ public void dfs(char[][] grid, int i, int j){
     dfs(grid,i,j + 1);
     dfs(grid,i,j - 1);
 }
+```
+```java
+//graph - dfs (和卡哥的代碼邏輯一致)
+class Solution {
+    boolean[][] visited;
+    int dir[][] = {
+        {0, 1}, //right
+        {1, 0}, //down
+        {-1, 0}, //up
+        {0, -1} //left
+    };
+    public int numIslands(char[][] grid) {
+        int count = 0;
+        visited = new boolean[grid.length][grid[0].length];
+
+        for(int i = 0; i < grid.length; i++){
+            for(int j = 0; j < grid[0].length; j++){
+                if(visited[i][j] == false && grid[i][j] == '1'){
+                    count++;
+                    dfs(grid, i, j);
+                }
+            }
+        }
+        return count;
+    }
+
+    private void dfs(char[][]grid, int x, int y){
+        if(visited[x][y] == true || grid[x][y] == '0')
+            return;
+	    
+        visited[x][y] = true;
+	
+        for(int i = 0; i < 4; i++){
+            int nextX = x + dir[i][0];
+            int nextY = y + dir[i][1];
+            if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length)
+                continue;
+            dfs(grid, nextX, nextY);
+        }
+    }
+}
+```
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