Merge pull request #1931 from StriveDD/master

添加0130.被围绕的区域的Java版本的 BFS、DFS 代码，0797.所有可能的路径的Java版本的 DFS 代码

problems/0130.被围绕的区域.md

@@ -84,7 +84,221 @@ public:
 
 ## 其他语言版本 
 
+### Java
+
+```Java
+// 广度优先遍历
+// 使用 visited 数组进行标记
+class Solution {
+    private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};  // 四个方向
+
+    public void solve(char[][] board) {
+        // rowSize：行的长度，colSize：列的长度
+        int rowSize = board.length, colSize = board[0].length; 
+        boolean[][] visited = new boolean[rowSize][colSize];
+        Queue<int[]> queue = new ArrayDeque<>();
+        // 从左侧边，和右侧边遍历
+        for (int row = 0; row < rowSize; row++) {
+            if (board[row][0] == 'O') {
+                visited[row][0] = true;
+                queue.add(new int[]{row, 0});
+            }
+            if (board[row][colSize - 1] == 'O') {
+                visited[row][colSize - 1] = true;
+                queue.add(new int[]{row, colSize - 1});
+            }
+        }
+        // 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
+        // 所以在遍历上边和下边时可以不用遍历四个角
+        for (int col = 1; col < colSize - 1; col++) {
+            if (board[0][col] == 'O') {
+                visited[0][col] = true;
+                queue.add(new int[]{0, col});
+            }
+            if (board[rowSize - 1][col] == 'O') {
+                visited[rowSize - 1][col] = true;
+                queue.add(new int[]{rowSize - 1, col});
+            }
+        }
+        // 广度优先遍历，把没有被 'X' 包围的 'O' 进行标记
+        while (!queue.isEmpty()) {
+            int[] current = queue.poll();
+            for (int[] pos: position) {
+                int row = current[0] + pos[0], col = current[1] + pos[1];
+                // 如果范围越界、位置已被访问过、该位置的值不是 'O'，就直接跳过
+                if (row < 0 || row >= rowSize || col < 0 || col >= colSize) continue;
+                if (visited[row][col] || board[row][col] != 'O') continue;
+                visited[row][col] = true;
+                queue.add(new int[]{row, col});
+            }
+        }
+        // 遍历数组，把没有被标记的 'O' 修改成 'X'
+        for (int row = 0; row < rowSize; row++) {
+            for (int col = 0; col < colSize; col++) {
+                if (board[row][col] == 'O' && !visited[row][col]) board[row][col] = 'X';
+            }
+        }
+    }
+}
+```
+```Java
+// 广度优先遍历
+// 直接修改 board 的值为其他特殊值
+class Solution {
+    private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};  // 四个方向
+
+    public void solve(char[][] board) {
+        // rowSize：行的长度，colSize：列的长度
+        int rowSize = board.length, colSize = board[0].length;
+        Queue<int[]> queue = new ArrayDeque<>();
+        // 从左侧边，和右侧边遍历
+        for (int row = 0; row < rowSize; row++) {
+            if (board[row][0] == 'O')
+                queue.add(new int[]{row, 0});
+            if (board[row][colSize - 1] == 'O')
+                queue.add(new int[]{row, colSize - 1});
+        }
+        // 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
+        // 所以在遍历上边和下边时可以不用遍历四个角
+        for (int col = 1; col < colSize - 1; col++) {
+            if (board[0][col] == 'O')
+                queue.add(new int[]{0, col});
+            if (board[rowSize - 1][col] == 'O')
+                queue.add(new int[]{rowSize - 1, col});
+        }
+        // 广度优先遍历，把没有被 'X' 包围的 'O' 修改成特殊值
+        while (!queue.isEmpty()) {
+            int[] current = queue.poll();
+            board[current[0]][current[1]] = 'A';
+            for (int[] pos: position) {
+                int row = current[0] + pos[0], col = current[1] + pos[1];
+                // 如果范围越界、该位置的值不是 'O'，就直接跳过
+                if (row < 0 || row >= rowSize || col < 0 || col >= colSize) continue;
+                if (board[row][col] != 'O') continue;
+                queue.add(new int[]{row, col});
+            }
+        }
+        // 遍历数组，把 'O' 修改成 'X'，特殊值修改成 'O'
+        for (int row = 0; row < rowSize; row++) {
+            for (int col = 0; col < colSize; col++) {
+                if (board[row][col] == 'A') board[row][col] = 'O';
+                else if (board[row][col] == 'O') board[row][col] = 'X'; 
+            }
+        }
+    }
+}
+```
+```Java
+// 深度优先遍历
+// 使用 visited 数组进行标记
+class Solution {
+    private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};  // 四个方向
+
+    public void dfs(char[][] board, int row, int col, boolean[][] visited) {
+        for (int[] pos: position) {
+            int nextRow = row + pos[0], nextCol = col + pos[1];
+            // 位置越界
+            if (nextRow < 0 || nextRow >= board.length || nextCol < 0 || nextCol >= board[0].length)
+                continue;
+            // 位置已被访问过、新位置值不是 'O'
+            if (visited[nextRow][nextCol] || board[nextRow][nextCol] != 'O') continue;
+            visited[nextRow][nextCol] = true;
+            dfs(board, nextRow, nextCol, visited);
+        }
+    }
+
+    public void solve(char[][] board) {
+        int rowSize = board.length, colSize = board[0].length;
+        boolean[][] visited = new boolean[rowSize][colSize];
+        // 从左侧遍、右侧遍遍历
+        for (int row = 0; row < rowSize; row++) {
+            if (board[row][0] == 'O' && !visited[row][0]) {
+                visited[row][0] = true;
+                dfs(board, row, 0, visited);
+            }
+            if (board[row][colSize - 1] == 'O' && !visited[row][colSize - 1]) {
+                visited[row][colSize - 1] = true;
+                dfs(board, row, colSize - 1, visited);
+            }
+        }
+        // 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
+        // 所以在遍历上边和下边时可以不用遍历四个角
+        for (int col = 1; col < colSize - 1; col++) {
+            if (board[0][col] == 'O' && !visited[0][col]) {
+                visited[0][col] = true;
+                dfs(board, 0, col, visited);
+            }
+            if (board[rowSize - 1][col] == 'O' && !visited[rowSize - 1][col]) {
+                visited[rowSize - 1][col] = true;
+                dfs(board, rowSize - 1, col, visited);
+            }
+        }
+        // 遍历数组，把没有被标记的 'O' 修改成 'X'
+        for (int row = 0; row < rowSize; row++) {
+            for (int col = 0; col < colSize; col++) {
+                if (board[row][col] == 'O' && !visited[row][col]) board[row][col] = 'X';
+            }
+        }
+    }
+}
+```
+```Java
+// 深度优先遍历
+// // 直接修改 board 的值为其他特殊值
+class Solution {
+    private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};  // 四个方向
+
+    public void dfs(char[][] board, int row, int col) {
+        for (int[] pos: position) {
+            int nextRow = row + pos[0], nextCol = col + pos[1];
+            // 位置越界
+            if (nextRow < 0 || nextRow >= board.length || nextCol < 0 || nextCol >= board[0].length)
+                continue;
+            // 新位置值不是 'O'
+            if (board[nextRow][nextCol] != 'O') continue;
+            board[nextRow][nextCol] = 'A';      // 修改为特殊值
+            dfs(board, nextRow, nextCol);
+        }
+    }
+
+    public void solve(char[][] board) {
+        int rowSize = board.length, colSize = board[0].length;
+        // 从左侧遍、右侧遍遍历
+        for (int row = 0; row < rowSize; row++) {
+            if (board[row][0] == 'O') {
+                board[row][0] = 'A';
+                dfs(board, row, 0);
+            }
+            if (board[row][colSize - 1] == 'O') {
+                board[row][colSize - 1] = 'A';
+                dfs(board, row, colSize - 1);
+            }
+        }
+        // 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
+        // 所以在遍历上边和下边时可以不用遍历四个角
+        for (int col = 1; col < colSize - 1; col++) {
+            if (board[0][col] == 'O') {
+                board[0][col] = 'A';
+                dfs(board, 0, col);
+            }
+            if (board[rowSize - 1][col] == 'O') {
+                board[rowSize - 1][col] = 'A';
+                dfs(board, rowSize - 1, col);
+            }
+        }
+        // 遍历数组，把 'O' 修改成 'X'，特殊值修改成 'O'
+        for (int row = 0; row < rowSize; row++) {
+            for (int col = 0; col < colSize; col++) {
+                if (board[row][col] == 'O') board[row][col] = 'X';
+                else if (board[row][col] == 'A') board[row][col] = 'O';
+            }
+        }
+    }
+}
+```
+
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
 </a>
+

problems/0797.所有可能的路径.md

@@ -161,6 +161,35 @@ public:
 
 ## Java 
 
+```Java
+// 深度优先遍历
+class Solution {
+    List<List<Integer>> ans;		// 用来存放满足条件的路径
+    List<Integer> cnt;		// 用来保存 dfs 过程中的节点值
+
+    public void dfs(int[][] graph, int node) {
+        if (node == graph.length - 1) {		// 如果当前节点是 n - 1，那么就保存这条路径
+            ans.add(new ArrayList<>(cnt));
+            return;
+        }
+        for (int index = 0; index < graph[node].length; index++) {
+            int nextNode = graph[node][index];
+            cnt.add(nextNode);
+            dfs(graph, nextNode);
+            cnt.remove(cnt.size() - 1);		// 回溯
+        }
+    }
+
+    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
+        ans = new ArrayList<>();
+        cnt = new ArrayList<>();
+        cnt.add(0);			// 注意，0 号节点要加入 cnt 数组中
+        dfs(graph, 0);
+        return ans;
+    }
+}
+```
+
 ## Python
 ```python
 class Solution:
@@ -192,3 +221,4 @@ class Solution:
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
 </a>
+