1 public void solve(char[][] board) {
2 if(board==null || board.length<=1 || board[0].length<=1)
3 return;
4
5 //第一行和最后一行进行fill
6 for(int i=0;i<board[0].length;i++){
7 fill(board,0,i);
8 fill(board,board.length-1,i);
9 }
10
11 //第一列和最后一列进行fill
12 for(int i=0;i<board.length;i++){
13 fill(board,i,0);
14 fill(board,i,board[0].length-1);
15 }
16
17 //对于当前格子中,所有O变成X(说明符合要求,没有被变成#),所有#变成O
18 for(int i=0;i<board.length;i++){
19 for(int j=0;j<board[0].length;j++){
20 if(board[i][j]==‘O‘)
21 board[i][j]=‘X‘;
22 else if(board[i][j]==‘#‘)
23 board[i][j]=‘O‘;
24 }
25 }
26 }
27
28 private void fill(char[][] board, int i, int j){
29 if(board[i][j]!=‘O‘)
30 return;
31 board[i][j] = ‘#‘;
32 //利用BFS
33 LinkedList<Integer> queue = new LinkedList<Integer>();
34 //先将矩阵的横纵坐标编码
35 int code = i*board[0].length+j;
36 queue.add(code);
37 while(!queue.isEmpty()){
38 code = queue.poll();
39 int row = code/board[0].length;//从code中还原横坐标
40 int col = code%board[0].length;//从code中还原纵坐标
41
42 if(row>=1 && board[row-1][col]==‘O‘){//当前元素上边是否为0
43 queue.add((row-1)*board[0].length + col);
44 board[row-1][col]=‘#‘;
45 }
46
47 if(row<=board.length-2 && board[row+1][col]==‘O‘){//当前元素下面是否为0
48 queue.add((row+1)*board[0].length + col);
49 board[row+1][col]=‘#‘;
50 }
51
52 if(col>=1 && board[row][col-1]==‘O‘){//当前元素左边是否为0
53 queue.add(row*board[0].length + col-1);
54 board[row][col-1]=‘#‘;
55 }
56
57 if(col<=board[0].length-2 && board[row][col+1]==‘O‘){//当前元素右边是否为0
58 queue.add(row*board[0].length + col+1);
59 board[row][col+1]=‘#‘;
60 }
61 }
62 }