Surrounded Regions .

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.

样例

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


代码如下，这次用的java因为c++其实并没有看过多少，只了解一点C的。
思路就是从边界的‘O’开始搜索与其相连的‘O’，并标记， 直到所有边界上的‘O’都被搜索过，没有被标记的就是被包围的。
本来想用BFS的，但是刚写到方法名想到其实不用bfs就能搞定，递归的出口也很明显，于是。。

public class Solution {
    /**
     * @param board a 2D board containing 'X' and 'O'
     * @return void
     */
    int flag[][]= new int[1000][1000];
    int x = 0;
    int y = 0;
    public void surroundedRegions(char[][] board) {
        // Write your code here
        this.x = board.length;
        if(board.length == 0) return ;
        this.y = board[0].length;
        for(int i = 0; i<x; i++) {
            if(board[i][0] == 'O') bfs(board,i,0);
            if(board[i][this.y-1] == 'O') bfs(board,i,this.y-1);
        }
        for(int i = 0; i<y ;i++) {
            if(board[0][i] == 'O') bfs(board,0,i);
             if(board[this.x-1][i] == 'O') bfs(board,this.x-1,i);
        }
        
        for(int i = 0; i < x; i++) {
            for(int j = 0; j < y; j++) {
                if(flag[i][j] == 0 && board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }
    
    public void bfs(char[][] board, int i, int j) {
        if(i<x && i>=0 && j<y && j>=0 && flag[i][j] == 0 && board[i][j] == 'O' ) {
            this.flag[i][j] = 1;
            bfs(board,i+1,j);
            bfs(board,i,j+1);
            bfs(board,i-1,j);
            bfs(board,i,j-1);
        }
    }
}