Add solution 130

Author: yanglbme

README.md

@@ -59,6 +59,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 092 | [Reverse Linked List II](https://github.com/doocs/leetcode/tree/master/solution/092.Reverse%20Linked%20List%20II) | `Linked List` |
 | 094 | [Binary Tree Inorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/094.Binary%20Tree%20Inorder%20Traversal) | `Hash Table`, `Stack`, `Tree` |
 | 127 | [Word Ladder](https://github.com/doocs/leetcode/tree/master/solution/127.Word%20Ladder) | `Breadth-first Search` |
+| 130 | [Surrounded Regions](https://github.com/doocs/leetcode/tree/master/solution/130.Surrounded%20Regions) | `Depth-first Search`, `Breadth-first Search`, `Union Find` |
 | 144 | [Binary Tree Preorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
 | 150 | [Evaluate Reverse Polish Notation](https://github.com/doocs/leetcode/tree/master/solution/150.Evaluate%20Reverse%20Polish%20Notation) | `Stack` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/doocs/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |

solution/130.Surrounded Regions/README.md

@@ -0,0 +1,139 @@
+## 被围绕的区域
+### 题目描述
+
+给定一个二维的矩阵，包含 `'X'` 和 `'O'`（**字母 O**）。
+
+找到所有被 `'X'` 围绕的区域，并将这些区域里所有的 `'O'` 用 `'X'` 填充。
+
+**示例:**
+```
+X X X X
+X O O X
+X X O X
+X O X X
+```
+
+运行你的函数后，矩阵变为：
+```
+X X X X
+X X X X
+X X X X
+X O X X
+```
+
+**解释:**
+
+被围绕的区间不会存在于边界上，换句话说，任何边界上的 `'O'` 都不会被填充为 `'X'`。 任何不在边界上，或不与边界上的 `'O'` 相连的 `'O'` 最终都会被填充为 `'X'`。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
+
+### 解法
+逆向思维，从上下左右四个边缘的有效点`'O'`往里搜索，将搜索到的有效点修改为`'Y'`。最后，剩下的`'O'`就是被`'X'`包围的，将它们修改为`'X'`，将`'Y'`修改回`'O'`。
+
+```java
+class Solution {
+
+    /**
+     * 坐标点
+     */
+    private class Point {
+        int x;
+        int y;
+
+        public Point(int x, int y) {
+            this.x = x;
+            this.y = y;
+        }
+    }
+
+    public void solve(char[][] board) {
+        if (board == null || board.length < 3 || board[0].length < 3) {
+            return;
+        }
+
+        int m = board.length;
+        int n = board[0].length;
+
+        // top & bottom
+        for (int i = 0; i < n; ++i) {
+            bfs(board, 0, i);
+            bfs(board, m - 1, i);
+        }
+
+        // left & right
+        for (int i = 1; i < m - 1; ++i) {
+            bfs(board, i, 0);
+            bfs(board, i, n - 1);
+        }
+
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (board[i][j] == 'O') {
+                    board[i][j] = 'X';
+                } else if (board[i][j] == 'Y') {
+                    board[i][j] = 'O';
+                }
+            }
+        }
+    }
+
+    /**
+     * 广度优先搜索
+     * @param board
+     * @param i
+     * @param j
+     */
+    private void bfs(char[][] board, int i, int j) {
+        Queue<Point> queue = new LinkedList<>();
+        if (isValid(board, i, j)) {
+            // 遇到'O'，修改为'Y'
+            board[i][j] = 'Y';
+            queue.offer(new Point(i, j));
+        }
+
+        while (!queue.isEmpty()) {
+            Point p = queue.poll();
+            // 获取下一层所有有效坐标点
+            List<Point> points = getNextLevelValidPoints(board, p.x, p.y);
+
+            for (Point point : points) {
+                queue.offer(point);
+            }
+        }
+
+    }
+
+    /**
+     * 获取下一层所有有效坐标点，将这些坐标点修改为 'Y' 并返回
+     * @param board
+     * @param i
+     * @param j
+     * @return list
+     */
+    private List<Point> getNextLevelValidPoints(char[][] board, int i, int j) {
+        List<Point> points = new ArrayList<>();
+        Point[] arr = new Point[] { new Point(i - 1, j), new Point(i + 1, j), new Point(i, j - 1),
+                new Point(i, j + 1) };
+        for (Point point : arr) {
+            if (isValid(board, point.x, point.y)) {
+                board[point.x][point.y] = 'Y';
+                points.add(point);
+            }
+        }
+        return points;
+    }
+
+    /**
+     * 判断坐标是否有效
+     * @param board
+     * @param i
+     * @param j
+     * @return boolean
+     */
+    private boolean isValid(char[][] board, int i, int j) {
+        int m = board.length;
+        int n = board[0].length;
+        // 当前坐标对应的值是'O'，才算有效
+        return (i < 0 || i > m - 1 || j < 0 || j > n - 1 || board[i][j] != 'O') ? false : true;
+    }
+
+}
+```

solution/130.Surrounded Regions/Solution.java

@@ -0,0 +1,110 @@
+class Solution {
+
+    /**
+     * 坐标点
+     */
+    private class Point {
+        int x;
+        int y;
+
+        public Point(int x, int y) {
+            this.x = x;
+            this.y = y;
+        }
+    }
+
+    public void solve(char[][] board) {
+        if (board == null || board.length < 3 || board[0].length < 3) {
+            return;
+        }
+
+        int m = board.length;
+        int n = board[0].length;
+
+        // top & bottom
+        for (int i = 0; i < n; ++i) {
+            bfs(board, 0, i);
+            bfs(board, m - 1, i);
+        }
+
+        // left & right
+        for (int i = 1; i < m - 1; ++i) {
+            bfs(board, i, 0);
+            bfs(board, i, n - 1);
+        }
+
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (board[i][j] == 'O') {
+                    board[i][j] = 'X';
+                } else if (board[i][j] == 'Y') {
+                    board[i][j] = 'O';
+                }
+            }
+        }
+    }
+
+    /**
+     * 广度优先搜索
+     * 
+     * @param board
+     * @param i
+     * @param j
+     */
+    private void bfs(char[][] board, int i, int j) {
+        Queue<Point> queue = new LinkedList<>();
+        if (isValid(board, i, j)) {
+            // 遇到'O'，修改为'Y'
+            board[i][j] = 'Y';
+            queue.offer(new Point(i, j));
+        }
+
+        while (!queue.isEmpty()) {
+            Point p = queue.poll();
+            // 获取下一层所有有效坐标点
+            List<Point> points = getNextLevelValidPoints(board, p.x, p.y);
+
+            for (Point point : points) {
+                queue.offer(point);
+            }
+        }
+
+    }
+
+    /**
+     * 获取下一层所有有效坐标点，将这些坐标点修改为 'Y' 并返回
+     * 
+     * @param board
+     * @param i
+     * @param j
+     * @return list
+     */
+    private List<Point> getNextLevelValidPoints(char[][] board, int i, int j) {
+        List<Point> points = new ArrayList<>();
+        Point[] arr = new Point[] { new Point(i - 1, j), new Point(i + 1, j), new Point(i, j - 1),
+                new Point(i, j + 1) };
+        for (Point point : arr) {
+            if (isValid(board, point.x, point.y)) {
+                board[point.x][point.y] = 'Y';
+                points.add(point);
+            }
+        }
+        return points;
+    }
+
+    /**
+     * 判断坐标是否有效
+     * 
+     * @param board
+     * @param i
+     * @param j
+     * @return boolean
+     */
+    private boolean isValid(char[][] board, int i, int j) {
+        int m = board.length;
+        int n = board[0].length;
+        // 当前坐标对应的值是'O'，才算有效
+        return (i < 0 || i > m - 1 || j < 0 || j > n - 1 || board[i][j] != 'O') ? false : true;
+    }
+
+}