leetcode130. Surrounded Regions

题目要求

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping 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

将所有被X环绕的O区域获取出来并转化为X

思路与代码

这篇题目与leetcode200. Number of Islands思路非常相近，建议毫无思路的同学先参考一下这篇博客。其实这种将区域相连的题目往往都可以使用深度优先遍历或者是Union-Find方法来实现。在这里我就给出深度优先遍历的实现方法，有兴趣的同学可以参考上文的博客来自己实现Union-Find方法。

和leetcode200题不同，在本题中，只有被完全包围的图形才可以被选中与转化，也就是，任何一个和数组边界上的O相连的都不可能是完全包围。所以在这题里我们采用逆向思维。将所有和边界O相连的O都标记出来。那么当我重新遍历数组的时候，剩下的O则是被完全包围的。

代码如下：

    public void solve(char[][] board) {
        if(board.length<=2 || board[0].length<=2 ) return;
        int row = board.length;
        int column = board[0].length;
        for(int i = 0 ; i<row ; i++){
            if(board[i][0] == 'O'){
                boundarySearch(board,i,0);
            }
            if(board[i][column-1]=='O'){
                boundarySearch(board, i, column-1);
            }
        }
        for(int j = 0 ; j<column ; j++){
            if(board[0][j] == 'O'){
                boundarySearch(board, 0, j);
            }
            if(board[row-1][j] == 'O'){
                boundarySearch(board, row-1, j);
            }
        }
        
        for(int i = 0 ; i<row ; i++){
            for(int j = 0 ; j<column ; j++){
                if(board[i][j]=='O'){
                    board[i][j] = 'X';
                }else if(board[i][j]=='*'){
                    board[i][j] = 'O';
                }
            }
        }
    }
    
    public void boundarySearch(char[][] board, int i, int j){
        if(i<0 || j<0 || i>=board.length || j>=board[0].length) return;
        if(board[i][j] == 'X' || board[i][j]=='*') return;
        board[i][j] = '*';
        if(i>1 && board[i-1][j]=='O') boundarySearch(board, i-1, j);
        if(j>1 && board[i][j-1]=='O') boundarySearch(board, i, j-1);
        if(i<board.length-2 && board[i+1][j]=='O') boundarySearch(board, i+1, j);
        if(j<board[0].length-2 && board[i][j+1]=='O') boundarySearch(board, i, j+1);
        
    }

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