130. Surrounded Regions

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本文详细介绍了如何通过BFS算法解决2D棋盘中被X包围的O问题，确保所有外围O都被正确转换为X。

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.

本题用DFS和BFS都可以（其实也都差不多）

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

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public class Solution {
public static void solve(char[][] board) {
    Queue<Integer> queue = new LinkedList<Integer>();
    if (board == null || board.length == 0)
      return;
    int m = board.length;
    int n = board[0].length;
    boolean visited[][] = new boolean[m][n];
    int dir[][] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {

        //以下是标准的BFS搜索，用visitedPoints记录访问的O
        if (board[i][j] == 'O' && !visited[i][j]) {
          boolean surounned = true;
          List<Integer> visitedPoints = new ArrayList<Integer>();
          queue.add(i * n + j);
          visited[i][j] = true;
          while (queue.size() > 0) {
            int point = queue.poll();
            visitedPoints.add(point);
            int x = point/n;
            int y = point%n;
            for (int k = 0; k < 4; k++) {
              int nextx = x + dir[k][0];
              int nexty = y + dir[k][1];
              if (nextx >= 0 && nextx < m && nexty >= 0 && nexty < n) {
                if (board[nextx][nexty] == 'O' && !visited[nextx][nexty])
                  queue.add(nextx * n + nexty);
                visited[nextx][nexty] = true;
              } else {
                surounned = false;
              }
            }
          }

          //如果当前遍历到的O是被包围的
          if (surounned) {
            for (int p : visitedPoints)
              board[p / n][p % n] = 'X';
          }
        }
      }
    }
  }
}