1. 题目

给定一个二维的矩阵，包含 ‘X’ 和 ‘O’（字母 O）。

找到所有被 ‘X’ 围绕的区域，并将这些区域里所有的 ‘O’ 用 ‘X’ 填充。

示例:
X X X X
X O O X
X X O X
X O X X
运行你的函数后，矩阵变为：X X X X
X X X X
X X X X
X O X X

解释:
任何边界上的 ‘O’ 都不会被填充为 ‘X’。 任何不在边界上，或不与边界上的 ‘O’ 相连的 ‘O’ 最终都会被填充为 ‘X’。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/surrounded-regions
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• 对四周的O进行BFS或者DFS，将其改为-
• 最后将O改成X，将-改成O

2.1 BFS

class Solution {int m, n;
public:void solve(vector<vector<char>>& board) {if(board.empty())return;m = board.size(), n = board[0].size();int i, j;for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){if((i==0 || i==m-1 || j == 0 || j == n-1) && board[i][j] == 'O')BFS(board,i,j);}}for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){if(board[i][j] == 'O')board[i][j] = 'X';else if(board[i][j] == '-')board[i][j] = 'O';}}}void BFS(vector<vector<char>>& board, int i, int j){queue<pair<int,int>> q;vector<vector<int>> dir = {{-1,0},{1,0},{0,1},{0,-1}};q.push({i,j});pair<int,int> tp;board[i][j] = '-';int k, x, y;while(!q.empty()){tp = q.front();q.pop();for(k = 0; k < 4; ++k){x = dir[k][0]+tp.first;y = dir[k][1]+tp.second;if(x>=0 && x<m && y>=0 && y<n && board[x][y]=='O'){q.push({x,y});board[x][y] = '-';}}}}
};

2.2 DFS

class Solution {int m, n;
public:void solve(vector<vector<char>>& board) {if(board.empty())return;m = board.size(), n = board[0].size();int i, j;vector<vector<int>> dir = {{-1,0},{1,0},{0,1},{0,-1}};for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){if((i==0 || i==m-1 || j == 0 || j == n-1) && board[i][j] == 'O')DFS(board,i,j,dir);}}for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){if(board[i][j] == 'O')board[i][j] = 'X';else if(board[i][j] == '-')board[i][j] = 'O';}}}void DFS(vector<vector<char>>& board, int i, int j, vector<vector<int>> &dir){board[i][j] = '-';int x, y;for(int k = 0; k < 4; ++k){x = dir[k][0]+i;y = dir[k][1]+j;if(x>=0 && x<m && y>=0 && y<n && board[x][y]=='O')DFS(board,x,y,dir);}}
};