lintcode477- Surrounded Regions- medium

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.

Example

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

After capture all regions surrounded by 'X', the board should be:

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

BFS。问题转换：一个0被x包围，等价于其不能连到地图外面。那如果对边缘一圈的O做BFS，碰到的O肯定都是能连到地图外的O，那就标记为‘W’(灌水)。最后再遍历一次地图，让所有被灌水的回到‘O’，其他的都归为‘X’即可。

数据结构：二维数组，统一一下函数都x,y传，数组调用都int[y][x]，x和w比y和h比，这样没问题。

细节：棋盘题进行上下左右移动可以借助int[] dx = {0, -1, 0, 1}; int[] dy = {-1, 0, 1, 0}; ，之后for(4) {newX = x + dx[i], newY = y + dy[i]}来做。

public class Solution {
    /*
     * @param board: board a 2D board containing 'X' and 'O'
     * @return: nothing
     */
    public void surroundedRegions(char[][] board) {
        // write your code here
        if (board.length == 0 || board[0].length == 0) {
            return;
        }
        int h = board.length;
        int w = board[0].length;
        
        for (int i = 0; i < w; i++) {
            bfs(board, 0, i);
            bfs(board, h - 1, i);
        }
        
        for (int j = 0; j < h; j++) {
            bfs(board, j, 0);
            bfs(board, j, w - 1);
        }
        
        for (int i = 0; i < w; i++) {
            for (int j = 0; j < h; j++) {
                if (board[j][i] == 'W') {
                    board[j][i] = 'O';
                } else {
                    board[j][i] = 'X';
                }
            }
        }
    }
    
    private void bfs(char[][] board, int sy, int sx) {
        
        if (board[sy][sx] == 'X') {
            return;
        }
        
        int h = board.length;
        int w = board[0].length;
        // 别写成0101
        int[] dx = {0, -1, 0, 1};
        int[] dy = {-1, 0, 1, 0};
        
        board[sy][sx] = 'W';
        for (int i = 0; i < 4; i++) {
            int cy = sy + dy[i];
            int cx = sx + dx[i];
            if (cx >= 0 && cx < w && cy >= 0 && cy < h && board[cy][cx] == 'O') {
                bfs(board, cy, cx);
            }
        }
    }
}