[LeetCode] Surrounded Regions

This problem is not quite difficult (a typical BFS traversal of graph), though, its aceptance rate is relatively low. In fact, the key obstacle in passing this problem is how to reduce the number of checks and avoid the annoying TLE.

There is a nice idea in this link. In fact, all the surrounded regions will not contain an O on the boundary. This idea takes advantage of this observation and visit the board from O's on the boundary and mark them using another character, say, #. Finally, all the remaining O's are the surrounded regions and should be captured to X. The # regions simply need to be recovered to O.

I adopt the idea from the above link. And I include some minor optimizations. For example, I pass the queue toVisit to the update function to check and update the four neighbors together during the BFS. This turns out to reduce the running time from 28ms to 16ms.

The code is as follows.

 1 class Solution {
 2 public:
 3     void solve(vector<vector<char>>& board) {
 4         if (board.empty()) return;
 5         int m = board.size(), n = board[0].size();
 6         for (int i = 0; i < m; i++) {
 7             if (board[i][0] == 'O')
 8                 mark(board, i, 0);
 9             if (board[i][n - 1] == 'O')
10                 mark(board, i, n - 1);
11         }
12         for (int j = 0; j < n; j++) {
13             if (board[0][j] == 'O')
14                 mark(board, 0, j);
15             if (board[m - 1][j] == 'O')
16                 mark(board, m - 1, j);
17         }
18         capture(board);
19     }
20 private:
21     // Update neighbors
22     void update(vector<vector<char>>& board, queue<pair<int, int>>& toVisit, int r, int c) {
23         int m = board.size(), n = board[0].size();
24         if (r - 1 >= 0 && board[r - 1][c] == 'O') {
25             board[r - 1][c] = '#';
26             toVisit.push(make_pair(r - 1, c));
27         }
28         if (r + 1 < m && board[r + 1][c] == 'O') {
29             board[r + 1][c] = '#';
30             toVisit.push(make_pair(r + 1, c));
31         }
32         if (c - 1 >= 0 && board[r][c - 1] == 'O') {
33             board[r][c - 1] = '#';
34             toVisit.push(make_pair(r, c - 1));
35         }
36         if (c + 1 < n && board[r][c + 1] == 'O') {
37             board[r][c + 1] = '#';
38             toVisit.push(make_pair(r, c + 1));
39         }
40     }
41     // Mark non-surrounded regions
42     void mark(vector<vector<char>>& board, int r, int c) {
43         queue<pair<int, int> > toVisit;
44         toVisit.push(make_pair(r, c));
45         board[r][c] = '#';
46         while (!toVisit.empty()) {
47             int num = toVisit.size();
48             for (int i = 0; i < num; i++) {
49                 pair<int, int> cur = toVisit.front();
50                 toVisit.pop();
51                 update(board, toVisit, cur.first, cur.second);
52             }
53         }
54     }
55     // Capture surrounded regions
56     void capture(vector<vector<char>>& board) {
57         int m = board.size(), n = board[0].size();
58         for (int i = 0; i < m; i++) {
59             for (int j = 0; j < n; j++) {
60                 if (board[i][j] == '#')
61                     board[i][j] = 'O';
62                 else board[i][j] = 'X';
63             }
64         }
65     }
66 };

 

转载于:https://www.cnblogs.com/jcliBlogger/p/4598456.html