【leetcode】Surrounded Regions

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

用dfs试试，大数据时递归会溢出

 1 class Solution {
 2 public:
 3     void solve(vector<vector<char>> &board) {
 4         int m=board.size();
 5         if(m==0) return;
 6         int n=board[0].size();
 7
 8         vector<vector<bool>> visited(m,vector<bool>(n,false));
 9         for(int i=0;i<m;i++)
10         {
11             if(board[i][0]==‘O‘&&!visited[i][0])
12             {
13                 dfs(board,visited,i,0,m,n);
14             }
15
16             if(board[i][n-1]==‘O‘&&!visited[i][n-1])
17             {
18                 dfs(board,visited,i,n-1,m,n);
19             }
20         }
21
22         for(int j=0;j<n;j++)
23         {
24
25             if(board[0][j]==‘O‘&&!visited[0][j])
26             {
27                 dfs(board,visited,0,j,m,n);
28             }
29
30             if(board[m-1][j]==‘O‘&&!visited[m-1][j])
31             {
32                 dfs(board,visited,m-1,j,m,n);
33             }
34
35         }
36
37
38         for(int i=0;i<m;i++)
39         {
40             for(int j=0;j<n;j++)
41             {
42                 if(board[i][j]==‘O‘&&visited[i][j])
43                 {
44                     board[i][j]=‘X‘;
45                 }
46             }
47         }
48
49         return;
50     }
51
52     void dfs(vector<vector<char>> &board,vector<vector<bool>> &visited,int i,int j,int &m,int &n)
53     {
54         if(board[i][j]==‘O‘)
55         {
56             visited[i][j]=true;
57             if(i+1<m&&visited[i+1][j]==false)dfs(board,visited,i+1,j,m,n);
58             if(j+1<n&&visited[i][j+1]==false)dfs(board,visited,i,j+1,m,n);
59             if(i-1>=0&&visited[i-1][j]==false)dfs(board,visited,i-1,j,m,n);
60             if(j-1>=0&&visited[i][j-1]==false)dfs(board,visited,i,j-1,m,n);
61         }
62         else
63         {
64             return;
65         }
66     }
67 };

利用广度优先搜索，先把边界上的O全部放到队列中，然后搜索

 1 class Solution {
 2 public:
 3     void solve(vector<vector<char>> &board) {
 4         int m=board.size();
 5         if(m==0) return;
 6         int n=board[0].size();
 7
 8         queue<pair<int,int>> q;
 9
10         vector<vector<bool>> visited(m,vector<bool>(n,false));
11         for(int i=0;i<m;i++)
12         {
13             if(board[i][0]==‘O‘) q.push(pair<int,int>(i,0));
14             if(board[i][n-1]==‘O‘)  q.push(pair<int,int>(i,n-1));
15         }
16
17         for(int j=0;j<n;j++)
18         {
19             if(board[0][j]==‘O‘)  q.push(pair<int,int>(0,j));
20             if(board[m-1][j]==‘O‘)  q.push(pair<int,int>(m-1,j));
21         }
22
23         bfs(q,board,visited,m,n);
24
25         for(int i=0;i<m;i++)
26         {
27             for(int j=0;j<n;j++)
28             {
29                 if(!visited[i][j]&&board[i][j]==‘O‘)  board[i][j]=‘X‘;
30             }
31         }
32     }
33
34
35     void bfs(queue<pair<int,int>> &q,vector<vector<char>> &board,vector<vector<bool>> &visited,int &m,int &n)
36     {
37         while(!q.empty())
38         {
39             int ii=q.front().first;
40             int jj=q.front().second;
41             visited[ii][jj]=true;
42
43             q.pop();
44
45             if(ii+1<m&&!visited[ii+1][jj]&&board[ii+1][jj]==‘O‘) q.push(pair<int,int>(ii+1,jj));
46             if(ii-1>=0&&!visited[ii-1][jj]&&board[ii-1][jj]==‘O‘) q.push(pair<int,int>(ii-1,jj));
47             if(jj+1<n&&!visited[ii][jj+1]&&board[ii][jj+1]==‘O‘) q.push(pair<int,int>(ii,jj+1));
48             if(jj-1>=0&&!visited[ii][jj-1]&&board[ii][jj-1]==‘O‘) q.push(pair<int,int>(ii,jj-1));
49         }
50     }
51 };