leetcode--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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

[java] view plain copy

1. public class Solution {  
2.     public void solve(char[][] board) {  
3.         int rows = board.length;  
4.         if(rows<=0) return;   
5.         int cols = board[0].length;  
6.         if(cols<=0) return;   
7.         boolean flag[][] = new boolean[rows][cols];  
8.         ArrayList<Integer> queue_x = new ArrayList<Integer>();  
9.         ArrayList<Integer> queue_y = new ArrayList<Integer>();  
10.         boolean OK = true;  
11.         int[] dir_x = new int[]{-1,1,0,0};  
12.         int[] dir_y = new int[]{0,0,-1,1};  
13.         for(int i=0;i<rows;i++){  
14.             for(int j=0;j<cols;j++){  
15.                 if(board[i][j]=='O'&&!flag[i][j]){                    
16.                     int low = 0;  
17.                     int high = 1;  
18.                     OK = true;  
19.                     queue_x.add(i);  
20.                     queue_y.add(j);                   
21.                     while(low<high){                                               
22.                         for(int k=0;k<4;k++){  
23.                             int x = queue_x.get(low)+dir_x[k];  
24.                             int y = queue_y.get(low)+dir_y[k];  
25.                             if(x>=0&&x<rows&&y>=0&&y<cols){  
26.                                 if(!flag[x][y]&&board[x][y]=='O'){  
27.                                     queue_x.add(x);  
28.                                     queue_y.add(y);  
29.                                     flag[x][y]=true;  
30.                                     high++;  
31.                                 }                                 
32.                             }else{  
33.                                 OK = false;  
34.                             }                                 
35.                         }                     
36.                         low++;  
37.                     }  
38.                     if(OK){  
39.                         for(int k=0;k<queue_x.size();k++){  
40.                             board[queue_x.get(k)][queue_y.get(k)] = 'X';  
41.                         }                         
42.                     }  
43.                     queue_x.clear();queue_y.clear();  
44.                 }  
45.             }  
46.         }         
47.     }  
48. }  

原文链接http://blog.csdn.net/crazy__chen/article/details/46524447