477 · 被围绕的区域

477 · 被围绕的区域

描述
给一个二维的矩阵，包含 ‘X’ 和 ‘O’, 找到所有被 ‘X’ 围绕的区域，并用 ‘X’ 替换其中所有的 ‘O’。

样例
样例 1:

输入:
X X X X
X O O X
X X O X
X O X X
输出:
X X X X
X X X X
X X X X
X O X X

public class Solution {
    /*
     * @param board: board a 2D board containing 'X' and 'O'
     * @return: nothing
     */

    int n , m  ;
    int No ;
    public void surroundedRegions(char[][] board) {
        // write your code here
        n = board.length ;
        if(n == 0){
            return  ;
        }
        if(n == 1){
            return ;
        }
        m = board[0].length ;
        int[][] mark = new int[n][m] ;
        No = 0 ;
        boolean[] surround = new boolean[n * m + 1] ;
        Arrays.fill(surround , true) ;
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < m ; j++){
                if(board[i][j] == 'O' && mark[i][j] == 0  ){
                    No++ ;
                    dfs(i , j , board , surround , mark ) ;
                }
            }
        }
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < m ; j++){     
                if(board[i][j] == 'O' && surround[mark[i][j]]){
                    board[i][j] = 'X' ;
                }
            }
        }

    }
    public void dfs(int sx , int sy ,char[][] board , boolean[] surround , int[][] mark ){
        if (sx < 0 || sy < 0 || sx >= n || sy >= m ) {
            return;              //特殊情况判断
        }
        int[] dx = {0 , 1 , 0 , -1} ;
        int[] dy = {1 , 0 , -1 , 0} ;
        Queue<Integer> qx = new LinkedList<>() ;
        Queue<Integer> qy = new LinkedList<>() ;
        qx.add(sx) ; 
        qy.add(sy) ;
        mark[sx][sy] = No ;
        while(!qx.isEmpty()){
            int ox = qx.poll() ;
            int oy = qy.poll() ;
            if(ox == 0 || ox == n-1 || oy == 0 || oy == m-1){
                surround[No] = false ;
            }
            for(int i = 0 ; i < 4 ; i++){
                int nx = ox + dx[i] ;
                int ny = oy + dy[i] ;
                if (nx < 0 || ny < 0 || nx >= n || ny >= m ) {
                   continue ;          //越界时
                 }
                if(0 <= ox && nx < n && 0 <= ny && ny < m && board[nx][ny] == 'O' && mark[nx][ny] == 0){
                    mark[nx][ny] =  No ;
                    qx.add(nx) ;
                    qy.add(ny) ;
                }
            }
        }

    }
}