Leetcode 130. Surrounded Regions

本文介绍了一种使用C++实现的棋盘翻转算法,该算法通过区域生长的方法来确定棋盘上哪些'O'可以被翻转为'X'。文章详细解释了算法的具体实现过程,并提供了一个具体的代码实现示例。
class Solution {
public:
    using TUPLE = tuple<int, int>;

    void solve(vector<vector<char>>& board) {

        int rows = board.size();
        if (rows == 0) return;
        int cols = board[0].size();

        vector<vector<int> > checkedTb(rows, vector<int>(cols, 0));

        for (int r = 0; r != rows; ++r) {
            for (int c = 0; c != cols; ++c) {
                if (board[r][c] == 'X' || checkedTb[r][c] == 1 ) continue;
                checkedTb[r][c] = 1;
                solver(board, make_tuple(r, c), checkedTb);
            }
        }
    }

private:
    // a region growed by point t
    void solver(vector<vector<char> >& board, TUPLE t, vector<vector<int> >& checkedTb) {
        int rows = board.size();
        int cols = board[0].size();

        bool open = (get<0>(t) == rows - 1 || get<0>(t) == 0 || get<1>(t) == cols - 1 || get<1>(t) == 0);


        // mark the points of regin with the mark of char '0';
        vector<TUPLE> points{ t };

        deque<TUPLE> dq{ t };

        while (!dq.empty()) {
            int n = dq.size();
            for (int i = 0; i != n; ++i) {
                TUPLE tp = dq.front();
                int r = get<0>(tp);
                int c = get<1>(tp);

                if (r != 0 && board[r - 1][c] == 'O' && checkedTb[r - 1][c] == 0) {
                    if (!open && r - 1 == 0)
                        open = true;
                    points.emplace_back(make_tuple(r - 1, c));
                    dq.emplace_back(make_tuple(r - 1, c));
                    checkedTb[r - 1][c] = 1;
                }

                if (r != rows - 1 && board[r + 1][c] == 'O' && checkedTb[r + 1][c] == 0) {
                    if (!open && r + 1 == rows - 1)
                        open = true;
                    points.emplace_back(make_tuple(r + 1, c));
                    dq.emplace_back(make_tuple(r + 1, c));
                    checkedTb[r + 1][c] = 1;
                }


                if (c != cols - 1 && board[r][c + 1] == 'O' && checkedTb[r][c + 1] == 0) {

                    if (!open && c + 1 == cols - 1)
                        open = true;
                    points.emplace_back(make_tuple(r, c + 1));
                    dq.emplace_back(make_tuple(r, c + 1));
                    checkedTb[r][c + 1] = 1;
                }

                if (c != 0 && board[r][c - 1] == 'O' && checkedTb[r][c - 1] ==  0) {

                    if (!open && c - 1 == 0)
                        open = true;
                    points.emplace_back(make_tuple(r, c - 1));
                    dq.emplace_back(make_tuple(r, c - 1));
                    checkedTb[r][c - 1] = 1;
                }

                dq.pop_front();
            }
        }

        if (!open) {
            for (auto & point : points) {
                int r = get<0>(point);
                int c = get<1>(point);
                board[r][c] = 'X';
            }
        }
        return;
    }
};
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