题目
Given a 2D board containing 'X' and 'O' (the letter 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
解题之法
class Solution {
public:
void solve(vector<vector<char> >& board) {
for (int i = 0; i < board.size(); ++i) {
for (int j = 0; j < board[i].size(); ++j) {
if ((i == 0 || i == board.size() - 1 || j == 0 || j == board[i].size() - 1) && board[i][j] == 'O')
solveDFS(board, i, j);
}
}
for (int i = 0; i < board.size(); ++i) {
for (int j = 0; j < board[i].size(); ++j) {
if (board[i][j] == 'O') board[i][j] = 'X';
if (board[i][j] == '$') board[i][j] = 'O';
}
}
}
void solveDFS(vector<vector<char> > &board, int i, int j) {
if (board[i][j] == 'O') {
board[i][j] = '$';
if (i > 0 && board[i - 1][j] == 'O')
solveDFS(board, i - 1, j);
if (j < board[i].size() - 1 && board[i][j + 1] == 'O')
solveDFS(board, i, j + 1);
if (i < board.size() - 1 && board[i + 1][j] == 'O')
solveDFS(board, i + 1, j);
if (j > 1 && board[i][j - 1] == 'O')
solveDFS(board, i, j - 1);
}
}
};
分析
扫描面矩阵的四条边，如果有O，则用DFS遍历，将所有连着的O都变成另一个字符，比如‘$’;这样剩下的O都是被包围的，然后将这些O变成X，再把'$'变回O就行了。
