leetcode-surrounded regions-ZZ

Problem Statement (link):

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

Analysis:
Rather than recording the 2D positions for any scanned ‘O’, a trick is to substitute any border ‘O’s with another character – here in the Code I use ‘Y’. And scan the board again to change any rest ‘O’s to ‘X’s, and change ‘Y’s back to ‘O’s.

We start searching ‘O’ from the four borders. I tried DFS first, the OJ gives Runtime error on the 250×250 large board; In the Sol 2 below, I implement BFS instead, and passed all tests.

The time complexity is O(n^2), as in the worst case, we may need to scan the entire board.

Code:
1, DFS

 class Solution {
 public:
     // dfs - Runtime error on large board 250x250
     void dfs(vector<vector<char>> &board, int r, int c) {
         if (r<||r>board.size()-||c<||c>board[].size()-||board[r][c]!='O')
             return;
         board[r][c]='Y';
         dfs(board, r-, c);
         dfs(board, r+, c);
         dfs(board, r, c-);
         dfs(board, r, c+);
     }
     void solve(vector<vector<char>> &board) {
         if (board.empty() || board.size()< || board[].size()<)
             return;
         int r=board.size();
         int c=board[].size();
         // dfs from boundary to inside
         for (int i=; i<c; i++) {
             if (board[][i]=='O')
                 dfs(board, , i);   // first row
             if (board[c-][i]=='O')
                 dfs(board, c-, i); // last row
         }
         for (int i=; i<board.size(); i++) {
             if (board[i][]=='O')
                 dfs(board, i, );   // first col
             if (board[i][c-])
                 dfs(board, i, c-); // last col
         }
         // scan entire matrix and set values
         for (int i=; i<board.size(); i++) {
             for (int j=; j<board[].size(); j++) {
                 if (board[i][j]=='O')
                     board[i][j]='X';
                 else if (board[i][j]=='Y')
                     board[i][j]='O';
             }
         }
     }
 };

2, BFS

 class Solution {
 public:
     void solve(vector<vector<char>> &board) {
         if (board.empty() || board.size()< || board[].size()<)
             return;
         int r=board.size();
         int c=board[].size();
         // queues to store row and col indices
         queue<int> qr;
         queue<int> qc;
         // start from boundary
         for (int i=; i<c; i++) {
             if (board[][i]=='O') { qr.push(); qc.push(i); }
             if (board[r-][i]=='O') { qr.push(r-); qc.push(i); }
         }
         for (int i=; i<r; i++) {
             if (board[i][]=='O') { qr.push(i); qc.push(); }
             if (board[i][c-]=='O') { qr.push(i); qc.push(c-); }
         }
         // BFS
         while (!qr.empty()) {
             int rt=qr.front(); qr.pop();
             int ct=qc.front(); qc.pop();
             board[rt][ct]='Y';
             if (rt->= && board[rt-][ct]=='O') { qr.push(rt-); qc.push(ct); }  //go up
             if (rt+<r && board[rt+][ct]=='O') { qr.push(rt+); qc.push(ct); } // go down
             if (ct->= && board[rt][ct-]=='O') { qr.push(rt); qc.push(ct-); } // go left
             if (ct+<c && board[rt][ct+]=='O') { qr.push(rt); qc.push(ct+); } // go right
         }         // scan entire matrix and set values
         for (int i=; i<board.size(); i++) {
             for (int j=; j<board[].size(); j++) {
                 if (board[i][j]=='O') board[i][j]='X';
                 else if (board[i][j]=='Y') board[i][j]='O';
             }
         }
     }
 };

http://justcodings.blogspot.com/2014/07/leetcode-surrounded-regions.html