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.

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

思路分析：这题比较简单，主要考察搜索，基本的思路是搜索和边缘的O连通的那些O，用特殊字符#标记出来，然后遍历矩阵把O赋为X，把#赋值为O。而搜索与边缘的O连通的O的方法，可以用DFS或者BFS，尝试从边缘每一个O开始搜索。但是这题用DFS递归写法会超时，因为测试用例有一个很大的矩阵，会导致递归栈溢出。下面AC Code用的是BFS，借助于队列实现的迭代式的BFS。另外队列保存的是矩阵中的数的一维index，要注意换算方法。这个算法在图形学中称为Flood fill算法。时间复杂度为O(mn)，空间复杂度为O（m+n）。

AC Code

public class Solution {
    //use bfs search(in an iterative manner with queue)
    public void bfs(char[][] board, int i, int j, int m, int n){
        board[i][j] = '#';
        //10:47
        //use queue to store all the adjacent node using sequcial index
        LinkedList queue = new LinkedList();
        int index = i*n + j;
        queue.add(index);
        while(!queue.isEmpty()){
            int cur = queue.poll();
            int row = cur / n;
            int col = cur % n;
            if(row+1 < m && board[row+1][col] == 'O'){
                queue.add((row+1)*n + col);
                board[row+1][col] = '#';
            }
            if(row-1 >= 0 && board[row-1][col] == 'O'){
                queue.add((row-1)*n + col);
                board[row-1][col] = '#';
            }
            if(col+1 < n && board[row][col+1] == 'O'){
                queue.add(row*n + col + 1);
                board[row][col+1] = '#';
            }
            if(col-1 >= 0 && board[row][col-1] == 'O'){
                queue.add(row*n + col - 1);
                board[row][col-1] = '#';
            }
        }
    }
    
    public void solve(char[][] board) {
        if(board == null) return;
        int m = board.length;
        if(m <= 1) return;
        int n = board[0].length;
        
        for(int i = 0; i < m; i++){
            if(board[i][0] == 'O'){
                bfs(board, i, 0, m, n);
            }
            if(board[i][n-1] == 'O'){
                bfs(board, i, n-1, m, n);
            }
        }
        for(int j = 0; j < n; j++){
            if(board[0][j] == 'O'){
                bfs(board, 0, j, m, n);
            }
            if(board[m-1][j] == 'O'){
                bfs(board, m-1, j, m, n);
            }
        }
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(board[i][j] == 'O'){
                    board[i][j] = 'X';   
                } else if(board[i][j] == '#'){
                    board[i][j] = 'O';
                }
            }
        }
    }
    
    //Use dfs search. If the input matrix is very large, it will stack overflow
    //So this problem should be solved by dfs in a iterative manner(with stack) or bfs in a iterative manner(with queue)
    //Could not be solved by dfs in recurcive way
    //0951
    /*public void dfs(char[][] board, int i, int j, int m, int n){
        board[i][j] = '#';
        if(i+1 < m && board[i+1][j] == 'O'){
            dfs(board, i+1, j, m, n);
        }
        if(i-1 >= 0 && board[i-1][j] == 'O'){
            dfs(board, i-1, j, m, n);
        }
        if(j+1 < n && board[i][j+1] == 'O'){
            dfs(board, i, j+1, m, n);
        }
        if(j-1 >= 0 && board[i][j-1] == 'O'){
            dfs(board, i, j-1, m, n);
        }
    }
    
    public void solve(char[][] board) {
        if(board == null) return;
        int m = board.length;
        if(m <= 1) return;
        int n = board[0].length;
        
        for(int i = 0; i < m; i++){
            if(board[i][0] == 'O'){
                dfs(board, i, 0, m, n);
            }
            if(board[i][n-1] == 'O'){
                dfs(board, i, n-1, m, n);
            }
        }
        for(int j = 0; j < n; j++){
            if(board[0][j] == 'O'){
                dfs(board, 0, j, m, n);
            }
            if(board[m-1][j] == 'O'){
                dfs(board, m-1, j, m, n);
            }
        }
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(board[i][j] == 'O'){
                    board[i][j] = 'X';   
                } else if(board[i][j] == '#'){
                    board[i][j] = 'O';
                }
            }
        }
    }*/
    //10:02
}