LeetCode: Surrounded Regions 解题报告

Surrounded Regions

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

SOLUTION 1:

经典 BFS 题目。如果使用DFS，会超时。

使用队列进行BFS。注意，在判断index越界时，主页君写了2种方法。

（1）可以直接对index在0-x*y之间取。这里有个小的trick，在点在左边缘的时候，index-1会导致我们回到上一行的最右边。当然那个点也是

边缘点，所以不会造成解的错误。

（2）判断点是不是在边缘，判定上下左右还有没有节点。

常出错的点：计算index是i * cols + j  这点要记好了。

 public class Solution {
     public void solve(char[][] board) {
         if (board == null || board.length == 0 || board[0].length == 0) {
             return;
         }

         int rows = board.length;
         int cols = board[0].length;

         // the first line and the last line.
         for (int j = 0; j < cols; j++) {
             bfs(board, 0, j);
             bfs(board, rows - 1, j);
         }

         // the left and right column
         for (int i = 0; i < rows; i++) {
             bfs(board, i, 0);
             bfs(board, i, cols - 1);
         }

         // capture all the nodes.
         for (int i = 0; i < rows; i++) {
             for (int j = 0; j < cols; j++) {
                 if (board[i][j] == 'O') {
                     board[i][j] = 'X';
                 } else if (board[i][j] == 'B') {
                     board[i][j] = 'O';
                 }
             }
         }

         return;
     }

     public void bfs1(char[][] board, int i, int j) {
         int rows = board.length;
         int cols = board[0].length;

         Queue<Integer> q = new LinkedList<Integer>();
         q.offer(i * cols + j);

         while (!q.isEmpty()) {
             int index = q.poll();

             // Index is out of bound.
             if (index < 0 || index >= rows * cols) {
                 continue;
             }

             int x = index / cols;
             int y = index % cols;

             if (board[x][y] != 'O') {
                 continue;
             }

             board[x][y] = 'B';
             q.offer(index + 1);
             q.offer(index - 1);
             q.offer(index + cols);
             q.offer(index - cols);
         }
     }

     public void bfs(char[][] board, int i, int j) {
         int rows = board.length;
         int cols = board[0].length;

         Queue<Integer> q = new LinkedList<Integer>();
         q.offer(i * cols + j);

         while (!q.isEmpty()) {
             int index = q.poll();

             int x = index / cols;
             int y = index % cols;

             if (board[x][y] != 'O') {
                 continue;
             }

             board[x][y] = 'B';
             if (y < cols - 1) {
                 q.offer(index + 1);
             }

             if (y > 0) {
                 q.offer(index - 1);
             }

             if (x > 0) {
                 q.offer(index - cols);
             }

             if (x < rows - 1) {
                 q.offer(index + cols);
             }
         }
     }
 }

SOLUTION 2:

附上DFS解法：

 public void dfs(char[][] board, int i, int j) {
         int rows = board.length;
         int cols = board[0].length;

         // out of bound or visited.
         if (i < 0 || i >= rows || j < 0 || j >= cols) {
             return;
         }

         if (board[i][j] != 'O') {
             return;
         }

         board[i][j] = 'B';

         // dfs the sorrounded regions.
         dfs(board, i + 1, j);
         dfs(board, i - 1, j);
         dfs(board, i, j + 1);
         dfs(board, i, j - 1);
     }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/bfs/Solve.java