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
思路：
反向思考，先找到没有被包围的区域，标记为‘+’,再把标为‘+’的区域标为‘O',标为’O'的区域改为‘X'。没有被包围的区域一定是与最外圈的’O'相连的区域，所以要先遍历区域的上下左右边界，找到‘O'的地方。接下来的问题是如何把所有与最外层‘O'相连的区域标记上。有两种思路：BFS和DFS。
所谓BFS是指把当前标记位置的上下左右都标记一遍，然后再标记相邻点的上下左右位置。不需递归，用队列。
DFS是指把当前标记位置的向一个方向标记，比如一直向左，直到没有可标记的，再换一个方向。需要递归。
代码里面BFS可以通过，DFS栈溢出了。
void solve(vector<vector<char>> &board) {
if(board.size() == ) return;
int rowNum = board.size();
int colNum = board[].size();
//遍历最外面一圈，找‘O'
//最上
for(int j = ; j < colNum; j++)
{
if(board[][j] == 'O')
BFS(board, , j);
}
//最下
for(int j = ; j < colNum; j++)
{
if(board[rowNum - ][j] == 'O')
BFS(board, rowNum - , j);
}
//最左
for(int i = ; i < rowNum; i++)
{
if(board[i][] == 'O')
BFS(board, i, );
}
//最右
for(int i = ; i < rowNum; i++)
{
if(board[i][colNum - ] == 'O')
BFS(board, i, colNum - );
}
for(int i = ; i < rowNum; i++)
{
for(int j = ; j < colNum; j++)
{
if(board[i][j] == 'O')
board[i][j] = 'X';
if(board[i][j] == '+')
board[i][j] = 'O';
}
}
}
void DFS(vector<vector<char>> &board, int r, int c)
{
if(r >= && c >= && r < board.size() && c < board[].size() && board[r][c] == 'O')
{
board[r][c] = '+';
DFS(board, r - , c);
DFS(board, r + , c);
DFS(board, r, c - );
DFS(board, r, c + );
}
}
void BFS(vector<vector<char>> &board, int r, int c)
{
queue<pair<int, int>> q;
q.push(make_pair(r, c));
while(!q.empty())
{
int i = q.front().first;
int j = q.front().second;
q.pop();
if(i >= && j >= && i < board.size() && j < board[].size() && board[i][j] == 'O')
{
board[i][j] = '+';
q.push(make_pair(i - , j));
q.push(make_pair(i + , j));
q.push(make_pair(i, j - ));
q.push(make_pair(i, j + ));
}
}
}
上面的代码已经是优化过的了，我自己写的时候只写出了DFS的，而且自己也没有意识到是DFS。代码也很繁琐。注意通过把判断条件放在一起来简化代码。
我原本很挫的代码：栈溢出。
class Solution {
public:
void solve(vector<vector<char>> &board) {
if(board.size() == ) return;
int rowNum = board.size();
int colNum = board[].size();
//遍历最外面一圈，找‘O'
//最上
for(int j = ; j < colNum; j++)
{
if(board[][j] == 'O')
{
board[][j] = '+';
mySolve(board, , j);
}
}
//最下
for(int j = ; j < colNum; j++)
{
if(board[rowNum - ][j] == 'O')
{
board[rowNum - ][j] = '+';
mySolve(board, rowNum - , j);
}
}
//最左
for(int i = ; i < rowNum; i++)
{
if(board[i][] == 'O')
{
board[i][] = '+';
mySolve(board, i, );
}
}
//最右
for(int i = ; i < rowNum; i++)
{
if(board[i][colNum - ] == 'O')
{
board[i][colNum - ] = '+';
mySolve(board, i, colNum - );
}
}
for(int i = ; i < rowNum; i++)
{
for(int j = ; j < colNum; j++)
{
if(board[i][j] == 'O')
board[i][j] = 'X';
if(board[i][j] == '+')
board[i][j] = 'O';
}
}
}
void mySolve(vector<vector<char>> &board, int r, int c)
{
if(r - >= && board[r - ][c] == 'O') //上
{
board[r - ][c] == '+';
mySolve(board, r - , c);
}
if(r + < board.size() && board[r + ][c] == 'O') //下
{
board[r + ][c] == '+';
mySolve(board, r + , c);
}
if(c - >= && board[r][c - ] == 'O') //左
{
board[r][c - ] == '+';
mySolve(board, r, c - );
}
if(c + < board[].size() && board[r][c + ] == 'O') //下
{
board[r][c + ] == '+';
mySolve(board, r, c + );
}
}
};