Surrounded-Regions

packagecom.ytx.array;

/**130题 

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

    A region is captured byflipping 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

 *

 *  给你一块2维的包含'X'或者'O'的板，找到所有被'X'包围的区域， 把区域内的‘O‘翻转成’X’。

 *@authoryuantian xin

 *

 *思路：连通性问题，用dfs和bfs都可以解决。逆向思维，求所有封闭的O，可以先求出所有连通的O，保存起来，再把其它的O改成X即可。只有在从边界上

 *连通的O才不会被X封闭。所以从边界上四条边界上进行深搜或者广搜，找到所有连通的O。

 */

publicclass Surrounded_regions {

       //二维矩阵的行数

       publicint rowNum;

       //二维矩阵的列数

       publicint colNum;

       

       /**

        * 利用深搜从边界上开始找，找到所有与边界上O连通的所有O,

        * 把所有连通的O变成*保存起来，那么剩下的O都是封闭的，被X包围的

        *@param index_x 二维数组的索引行

        *@param index_y 二维数据的索引列

        *@param board   二维数据

        */

       publicvoid dfs(intindex_x,int index_y, char board[][]) {

             if(board[index_x][index_y] == 'O') {

                    //是O才继续往下找，把所有连通的O都变成*，之后再变回来

                    board[index_x][index_y] = '*';

                    

                    //向上找

                    if(index_x> 1) {

                           dfs(index_x- 1,index_y,board);

                    }

                    

                    //向下找

                    if(index_x< rowNum - 1) {

                           dfs(index_x+1,index_y,board);

                    }

                    

                    //向左找

                    if(index_y> 1) {

                           dfs(index_x,index_y - 1, board);

                    }

                    

                    //向右找

                    if(index_y< colNum -1) {

                           dfs(index_x,index_y + 1, board);

                    }

             }

       }

       

       /**

        * 解决的办法，找到所有与边界上的O连通的O，改成*，剩下的不连通的

        * 改成X，最后把*变回O即可

        *@param board

        */

       publicvoid solve(char[][]board) {

             

             if(board== null || board.length<= 0 || board[0].length<= 0) {

                    return;

             }

             rowNum= board.length;

             colNum= board.length;

             

             //从第0列和最后一列开始找，递归深搜

             for(inti = 0; i < rowNum;i++) {

                    dfs(i,0,board);

                    dfs(i,colNum- 1, board);

             }

             

             //从第一行和最后一行开始找

             for(intj = 0; j < colNum;j++) {

                    dfs(0,j,board);

                    dfs(rowNum- 1, j, board);

             }

             

             //遍历二维数组，剩下不连通的O全部变成X,把*变回O。

             for(inti = 0; i < rowNum;i++) {

                    for(intj = 0; j < colNum;j++) {

                           

                           if(board[i][j] == 'O') {

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

                           }

                           

                           if(board[i][j] == '*') {

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

                           }

                    }

             }

       }

       publicstatic void main(String[]args) {

             charboard [][] ={{'X','X','X','X'},{'X','O','O','X'},{'X','X','O','X'},{'X','O','X','X'}};

             for(inti = 0; i < board.length;i++) {

                    for(intj = 0; j < board[0].length;j++) {

                           

                           System.out.print(board[i][j] + " ");

                    }

                    System.out.println("");

             }

             

             System.out.println("---------");

             Surrounded_regionssr = newSurrounded_regions();

             sr.solve(board);

             for(inti = 0; i < board.length;i++) {

                    for(intj = 0; j < board[0].length;j++) {

                           

                           System.out.print(board[i][j] + " ");

                    }

                    System.out.println("");

             }

       }

}