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【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 }