class Solution {
public:
void solve(vector<vector<char>>& board) {
if (board.size() <= 0 || board[0].size() <= 0)
return;
int row = board.size();
int col = board[0].size();
// 第一列和最后一列进行DFS,将与边界O相邻的O全部标记为*
for (int i = 0; i < row; ++i){
if (board[i][0] == 'O')
search(board, i, 0);
if (board[i][col - 1] == 'O')
search(board, i, col - 1);
}
for (int i = 0; i < col; ++i){
if (board[0][i] == 'O')
search(board, 0 , i);
if (board[row - 1][i] == 'O')
search(board, row - 1, i);
}
// 最后,标记为‘*’的表示,这块O连通图为没有被包围的,还原为'O',仍然为‘O’的表示被X包围了,要改成‘X’
for (int i = 0; i < row; ++i)
for (int j = 0; j < col; ++j)
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == '*')
board[i][j] = 'O';
}
void search(vector<vector<char>>& board, int row, int col){
if (row < 0 || col < 0 || row >= board.size() || col >= board[0].size())
return;
if (board[row][col] != 'O')
return;
board[row][col] = '*';
search(board , row + 1, col);
search(board , row - 1, col);
search(board , row , col + 1);
search(board , row , col - 1);
}
};
static int x=[](){
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return 0;
}();
LetCode 130. 被围绕的区域
猜你喜欢
转载自blog.csdn.net/wbb1997/article/details/81123593
今日推荐
周排行