@@ -142,6 +142,53 @@ class Solution {
142142 return landSum * 4 - cover * 2 ;
143143 }
144144}
145+ // 延伸 - 傳統DFS解法(使用visited數組)(遇到邊界 或是 海水 就edge ++)
146+ class Solution {
147+ int dir[][] = {
148+ {0 , 1 },
149+ {0 , - 1 },
150+ {1 , 0 },
151+ {- 1 , 0 }
152+ };
153+ 154+ boolean visited[][];
155+ int res = 0 ;
156+ 157+ public int islandPerimeter (int [][] grid ) {
158+ int row = grid. length;
159+ int col = grid[0 ]. length;
160+ visited = new boolean [row][col];
161+ 162+ int result = 0 ;
163+ 164+ for (int i = 0 ; i < row; i++ ){
165+ for (int j = 0 ; j < col; j++ ){
166+ if (visited[i][j] == false && grid[i][j] == 1 )
167+ result += dfs(grid, i, j);
168+ }
169+ }
170+ return result;
171+ }
172+ 173+ private int dfs (int [][] grid , int x , int y ){
174+ // 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)
175+ if (x < 0 || y < 0 || x >= grid. length || y >= grid[0 ]. length || grid[x][y] == 0 )
176+ return 1 ;
177+ // 如果該地已經拜訪過,就return 0 避免重複計算
178+ if (visited[x][y])
179+ return 0 ;
180+ int temp = 0 ;
181+ visited[x][y] = true ;
182+ for (int i = 0 ; i < 4 ; i++ ){
183+ int nextX = x + dir[i][0 ];
184+ int nextY = y + dir[i][1 ];
185+ // 用temp 把edge存起來
186+ temp += dfs(grid, nextX, nextY);
187+ }
188+ return temp;
189+ }
190+ }
191+ 145192```
146193
147194Python:
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