【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：

 1 public class Solution {

 2     class co{

 3         int x;

 4         int y;

 5         public co(int x,int y){

 6             this.x = x;

 7             this.y = y;

 8         }

 9     }

10     public void solve(char[][] board) {

11         if(board == null || board.length == 0)

12             return;

13         int m = board.length;

14         int n = board[0].length;

15         boolean[][] visited = new boolean[m][n];

16         boolean[][] isX = new boolean[m][n];

17         for(boolean[] row:isX)

18             Arrays.fill(row, true);

19         //put all o's cordinates into queue

20         Queue<co> queue = new LinkedList<co>();

21         

22         //first line and last line

23         for(int i  = 0;i < n;i++)

24         {

25             if(board[0][i]=='O'){

26                 queue.add(new co(0, i));

27                 visited[0][i]= true; 

28                 isX[0][i]= false; 

29             }

30             if(board[m-1][i]=='O'){

31                 queue.add(new co(m-1, i));

32                 visited[m-1][i]= true; 

33                 isX[m-1][i]= false;

34             }

35         }

36         

37         //first and last column

38         for(int j = 0;j<m;j++){

39             if(board[j][0]=='O'){

40                 queue.add(new co(j, 0));

41                 visited[j][0]= true; 

42                 isX[j][0]= false;

43             }

44             if(board[j][n-1]=='O'){

45                 queue.add(new co(j,n-1));

46                 visited[j][n-1]= true; 

47                 isX[j][n-1]= false;

48             }

49         }

50         

51         while(!queue.isEmpty()){

52             co c = queue.poll();

53             //up

54             if(c.x >= 1 && board[c.x-1][c.y] == 'O'&&!visited[c.x-1][c.y]){

55                 visited[c.x-1][c.y] = true;

56                 isX[c.x-1][c.y] = false;

57                 queue.add(new co(c.x-1, c.y));

58             }

59             //down

60             if(c.x+1<m && board[c.x+1][c.y]=='O' && !visited[c.x+1][c.y]){

61                 visited[c.x+1][c.y] = true;

62                 isX[c.x+1][c.y]= false; 

63                 queue.add(new co(c.x+1, c.y));

64             }

65             //left

66             if(c.y-1>=0 && board[c.x][c.y-1]=='O' && !visited[c.x][c.y-1]){

67                 visited[c.x][c.y-1] = true;

68                 isX[c.x][c.y-1] = false;

69                 queue.add(new co(c.x, c.y-1));

70             }

71             //right

72             if(c.y+1<n && board[c.x][c.y+1] == 'O' && !visited[c.x][c.y+1]){

73                 visited[c.x][c.y+1] = true;

74                 isX[c.x][c.y+1] = false;

75                 queue.add(new co(c.x, c.y+1));

76             }            

77         }

78         for(int i = 0;i < m;i++){

79             for(int j = 0;j < n;j++){

80                 if(isX[i][j] )

81                     board[i][j]= 'X'; 

82             }

83         }

84         return;

85     }

86 }