Surrounded Regions

https://oj.leetcode.com/problems/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.

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

这样的题目要得到一个答案，其实DFS和BFS都是可行的，理论上也不存在实际performance上的差距。本质的做法就是对每一片联通的0区域，做一次DFS或者BFS遍历其有效性并且改变其值。为了对联通区域只做一次遍历，标记是必须的。而不管是BFS还是DFS标记的方式一般来说是用HashMap或者HashSet记录路径，但是这一题的特质表示它其实可以in-place。把遍历过的O直接变成X（成功的情况）或者T（不成功的情况）即可，这样就标记了。最后把标记为T的改回O就好。这样就能inplace了。

这一题leetcode的online judge对performance要求有点奇怪，虽然算法角度上，DFS的实际performance不会比BFS差（甚至更高，DFS递归只需要每个元素access不超过两次）。但事实上oj上放了一个很极端的case会让DFS出现stackoverflow的exception，也就是递归层级爆掉了。所以下面先放BFS的代码：

    public void solve(char[][] board) { 
        if(board.length == 0 || board[0].length == 0) 
            return; 
        for(int i = 1; i < board.length - 1; i++){ 
            for(int j = 1; j < board[0].length - 1; j++){ 
                if(board[i][j] == 'O')helper(board, i, j); 
            } 
        } 
        for(int i = 0; i < board.length; i++) 
            for(int j = 0; j < board[0].length; j++) 
                board[i][j] = board[i][j] == 'T' ? 'O' : board[i][j]; 
    } 
     
    public void helper(char[][] board, int i, int j){ 
        boolean isValid = true; 
        Queue<Point> queue = new LinkedList<Point>(); 
        LinkedList<Point> res_candidate = new LinkedList<Point>(); 
        queue.add(new Point(i, j)); 
        while(!queue.isEmpty()){ 
            Point p = queue.poll(); 
            int x = p.x; 
            int y = p.y; 
            if(board[x][y] == 'T' || board[x][y] == 'X') 
                continue; 
            if(x == 0 || y == 0 || x == board.length - 1 || y == board[0].length - 1){ 
                isValid = false; 
                continue; 
            } 
            board[x][y] = 'T'; 
            res_candidate.add(p); 
            queue.add(new Point(x - 1, y)); 
            queue.add(new Point(x + 1, y)); 
            queue.add(new Point(x, y - 1)); 
            queue.add(new Point(x, y + 1)); 
        } 
        if(isValid){ 
            for(Point p : res_candidate) 
                board[p.x][p.y] = 'X'; 
        } 
    }

其实可以不用那个list，如果判断为true再跑一次BFS就好了。只是code写起来真的很无聊，所以就省事了。下面放一个会爆stack的DFS，感觉还是值得参考一下的：

    public void solve(char[][] board) {
        if(board.length == 0 || board[0].length == 0)
            return;
        for(int i = 1; i < board.length - 1; i++){
            for(int j = 1; j < board[0].length - 1; j++){
                if(board[i][j] == 'O' && helper(board, i, j));
            }
        }
        for(int i = 0; i < board.length; i++)
            for(int j = 0; j < board[0].length; j++)
                board[i][j] = board[i][j] == 'T' ? 'O' : board[i][j];
    }
    
    public boolean helper(char[][] board, int i, int j){
        if(board[i][j] == 'X' || board[i][j] == 'T')
            return true;
        if(i == 0 || i == board.length - 1 || j == 0 || j == board[0].length - 1)
            return false;
        board[i][j] = 'T';
        if(helper(board, i - 1, j) && helper(board, i + 1, j) && helper(board, i, j - 1) && helper(board, i, j + 1)){
            board[i][j] = 'X';
            return true;
        }else
            return false;
    }