# [Leetcode questions and Notes] surrounded regions

Source: Internet
Author: User

Given a 2D Board containing`‘X‘`And`‘O‘`, Capture all regions surrounded`‘X‘`.

A region is captured by flipping all`‘O‘`S`‘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 shoshould be:

`X X X XX X X XX X X XX O X X`

Problem: BFS is used.

The peripheral o cannot be changed to X, and these o can be pushed into a queue, and then the breadth is first searched from top to bottom and left, the O around them does not need to be changed to X (because it has a path of o that breaks out), and then continues to search for the o from the surrounding points ......

Implementation Details:

• Use a visited two-dimensional array to record whether an O has been in the queue, so that it does not have an endless loop;
• In another two-dimensional array, ISX records vertices that do not need to be converted to X. After BFS ends, all vertices to be converted to X are converted to X;
• In the question, both X and O are uppercase =. =

The implementation code is as follows:

` 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 queue20         Queue<co> queue = new LinkedList<co>();21         22         //first line and last line23         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 column38         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             //up54             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             //down60             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             //left66             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             //right72             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 }`

[Leetcode questions and Notes] surrounded regions

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