diff --git a/problems/0463.岛屿的周长.md b/problems/0463.岛屿的周长.md
index 2f399f40..18f1d01e 100644
--- a/problems/0463.岛屿的周长.md
+++ b/problems/0463.岛屿的周长.md
@@ -142,6 +142,53 @@ class Solution {
         return landSum * 4 - cover * 2;
     }
 }
+// 延伸 - 傳統DFS解法(使用visited數組)（遇到邊界 或是 海水 就edge ++）
+class Solution {
+    int dir[][] ={
+        {0, 1},
+        {0, -1},
+        {1, 0},
+        {-1, 0}
+    };
+
+    boolean visited[][];
+    int res = 0;
+
+    public int islandPerimeter(int[][] grid) {
+        int row = grid.length;
+        int col = grid[0].length;
+        visited = new boolean[row][col];
+
+        int result = 0;
+        
+        for(int i = 0; i < row; i++){
+            for(int j = 0; j < col; j++){
+                if(visited[i][j] == false && grid[i][j] == 1)
+                    result += dfs(grid, i, j);
+            }
+        }
+        return result;
+    }
+
+    private int dfs(int[][] grid, int x, int y){
+        //如果遇到 邊界（x < 0 || y < 0 || x >= grid.length || y >= grid[0].length）或是 遇到海水(grid[x][y] == 0)就return 1（edge + 1）
+        if(x < 0 || y < 0 || x >= grid.length || y >= grid[0].length || grid[x][y] == 0)
+            return 1;
+        //如果該地已經拜訪過，就return 0 避免重複計算
+        if(visited[x][y])
+            return 0;
+        int temp = 0;
+        visited[x][y] = true;
+        for(int i = 0; i < 4; i++){
+            int nextX = x + dir[i][0];
+            int nextY = y + dir[i][1];
+            //用temp 把edge存起來
+            temp +=dfs(grid, nextX, nextY);
+        }
+        return temp;
+    }
+}
+
 ```
 
 Python：