【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;
     }
 }