【leetcode刷题笔记】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

---

题解：用的BFS。

    最外围的O肯定是没法变成X的，把这些O推进一个队列里面，然后从它们开始上下左右进行广度优先搜索，它们周围的O也是不用变成X的（因为它有一条突围出去的都是O的路径），然后从周围的点继续广搜......

实现细节：

• 用一个visited二维数组记录某个O是否已经进过队列了，这样就不会死循环；
• 另一个二维数组isX记录哪些点不用变成X，在BFS结束后，把要变成X的点都变成X；
• 题目中X和O都是大写的=。=

实现代码如下：

public class Solution {
    class co{
        int x;
        int y;
        public co(int x,int y){
            this.x = x;
            this.y = y;
        }
    }
    public void solve(char[][] board) {
        if(board == null || board.length == 0)
            return;
        int m = board.length;
        int n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        boolean[][] isX = new boolean[m][n];
        for(boolean[] row:isX)
            Arrays.fill(row, true);
        //put all o's cordinates into queue
        Queue<co> queue = new LinkedList<co>();
        
        //first line and last line
        for(int i  = 0;i < n;i++)
        {
            if(board[0][i]=='O'){
                queue.add(new co(0, i));
                visited[0][i]= true; 
                isX[0][i]= false; 
            }
            if(board[m-1][i]=='O'){
                queue.add(new co(m-1, i));
                visited[m-1][i]= true; 
                isX[m-1][i]= false;
            }
        }
        
        //first and last column
        for(int j = 0;j<m;j++){
            if(board[j][0]=='O'){
                queue.add(new co(j, 0));
                visited[j][0]= true; 
                isX[j][0]= false;
            }
            if(board[j][n-1]=='O'){
                queue.add(new co(j,n-1));
                visited[j][n-1]= true; 
                isX[j][n-1]= false;
            }
        }
        
        while(!queue.isEmpty()){
            co c = queue.poll();
            //up
            if(c.x >= 1 && board[c.x-1][c.y] == 'O'&&!visited[c.x-1][c.y]){
                visited[c.x-1][c.y] = true;
                isX[c.x-1][c.y] = false;
                queue.add(new co(c.x-1, c.y));
            }
            //down
            if(c.x+1<m && board[c.x+1][c.y]=='O' && !visited[c.x+1][c.y]){
                visited[c.x+1][c.y] = true;
                isX[c.x+1][c.y]= false; 
                queue.add(new co(c.x+1, c.y));
            }
            //left
            if(c.y-1>=0 && board[c.x][c.y-1]=='O' && !visited[c.x][c.y-1]){
                visited[c.x][c.y-1] = true;
                isX[c.x][c.y-1] = false;
                queue.add(new co(c.x, c.y-1));
            }
            //right
            if(c.y+1<n && board[c.x][c.y+1] == 'O' && !visited[c.x][c.y+1]){
                visited[c.x][c.y+1] = true;
                isX[c.x][c.y+1] = false;
                queue.add(new co(c.x, c.y+1));
            }            
        }
        for(int i = 0;i < m;i++){
            for(int j = 0;j < n;j++){
                if(isX[i][j] )
                    board[i][j]= 'X'; 
            }
        }
        return;
    }
}