LeetCode *** 130. Surrounded Regions

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题目：

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

分析：

代码：

class Solution {
public:int m, n;void solve(vector<vector<char>>& board) {m = board.size();if (m == 0 || m == 1)return;n = board[0].size();if (n == 1)return;queue<pair<int, int>> record;for (int j = 0; j<n; ++j) {if (board[0][j] == 'O')record.push(pair<int, int>(0, j));if (board[m - 1][j] == 'O')record.push(pair<int, int>(m - 1, j));}for (int i = 0; i<m; ++i) {if (board[i][0] == 'O')record.push(pair<int, int>(i, 0));if (board[i][n - 1] == 'O')record.push(pair<int, int>(i, n - 1));}while (!record.empty()) {pair<int, int> tmp = record.front();record.pop();int i = tmp.first, j = tmp.second;board[tmp.first][tmp.second] = 'T';if (i>0 && board[i - 1][j] == 'O')record.push(pair<int,int>(i - 1, j));if (i<m - 1 && board[i + 1][j] == 'O')record.push(pair<int,int>(i + 1, j));if (j>0 && board[i][j - 1] == 'O')record.push(pair<int,int>(i, j - 1));if (j<n - 1 && board[i][j + 1] == 'O')record.push(pair<int,int>(i, j + 1));}for (int i = 0; i<m; ++i)for (int j = 0; j<n; ++j)if (board[i][j] == 'O')board[i][j] = 'X';else if (board[i][j] == 'T')board[i][j] = 'O';}
};

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