LeetCode 0130 - Surrounded Regions

Reck Zhang

发布于 2021-08-11 14:36:39

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发布于 2021-08-11 14:36:39

文章被收录于专栏：Reck ZhangReck Zhang

Surrounded Regions

Desicription

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.

Solution

class Solution {
private:
    void bfs(int x, int y, vector<vector<char>>& board) {
        int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
        bool flag = 0;
        queue<pair<int, int>> path;
        vector<pair<int, int>> book;
        path.push(make_pair(x, y));
        book.push_back(make_pair(x, y));
        board[x][y] = 'X';
        if(x == 0 || x == board.size()-1 || y == 0 || y == board[0].size()-1)
            flag = 1;
        while(path.size()) {
            int currentX = path.front().first;
            int currentY = path.front().second;
            path.pop();
            for(int i = 0; i < 4; i++) {
                int dx = currentX + dir[i][0];
                int dy = currentY + dir[i][1];
                if(dx < 0 || dx >= board.size() || dy < 0 || dy >= board[0].size() || board[dx][dy] != 'O')
                    continue;
                path.push(make_pair(dx, dy));
                book.push_back(make_pair(dx, dy));
                board[dx][dy] = 'X';
                if(dx == 0 || dx == board.size()-1 || dy == 0 || dy == board[0].size()-1)
                    flag = 1;
            }
        }
        if(flag) {
            for(auto& it : book)
                board[it.first][it.second] = 'N';
        }
    }
public:
    void solve(vector<vector<char>>& board) {
        for(int i = 0; i < board.size(); i++) {
            for(int j = 0 ; j < board[0].size(); j++) {
                if(board[i][j] == 'O') {
                    bfs(i, j, board);
                }
            }
        }
        for(int i = 0; i < board.size(); i++) {
            for(int j = 0 ; j < board[0].size(); j++) {
                if(board[i][j] == 'N') {
                    board[i][j] = 'O';
                }
            }
        }
    }
};

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原始发表：2018-05-24，如有侵权请联系 cloudcommunity@tencent.com 删除

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