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130. Surrounded Regions.cpp
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130. Surrounded Regions.cpp
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// ***
//
// 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.
//
// 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
//
// Explanation:
// Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to
// 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two
// cells are connected if they are adjacent cells connected horizontally or vertically.
//
// ***
//
// First traverse the four edges for 'O', DFS the 'O's and change all 'O' to '$'.
// Now, all 'O' left (those that have not been dfs-ed) must have been surrounded by 'X'. Change all of them to 'X'.
// Then, change the '$'s back to 'O'.
class Solution {
public:
void solve(vector<vector<char>>& board) {
// Traverse four edges and change all O to $
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') {
_dfs(board, i, j);
}
}
}
// Flip remaining O to X and flip $ back to O
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] == '$') {
board[i][j] = 'O';
}
}
}
}
private:
void _dfs(vector<vector<char>>& board, int i, int j) {
if (i < 0 || j < 0 || i >= board.size() || j >= board[0].size() || board[i][j] != 'O') {
return;
}
// Turn all 'O' to '$'.
board[i][j] = '$';
_dfs(board, i + 1, j);
_dfs(board, i - 1, j);
_dfs(board, i, j + 1);
_dfs(board, i, j - 1);
}
};