Time Cost

26min45s

Implementation

First push all ‘O’ on edge in a queue. Then use DFS to deal with O in the queue.

Code

• Solution

  class Solution {
  public:
    void solve(vector<vector<char>>& board) {
        int m = board.size();
        if (m == 0) return;
        int n = board[0].size();
        if (n == 0) return;
  
        queue<pair<int,int>> q;
  
        auto push_if_O = [&](int x, int y) {
            if (board[x][y] == 'O') {
                board[x][y] = 'E';         
                q.push({x, y});
            }
        };
  
        for (int i = 0; i < m; i++) {
            push_if_O(i, 0);
            push_if_O(i, n - 1);
        }
        for (int j = 0; j < n; j++) {
            push_if_O(0, j);
            push_if_O(m - 1, j);
        }
  
        static const int dx[4] = {1, -1, 0, 0};
        static const int dy[4] = {0, 0, 1, -1};
  
        while (!q.empty()) {
            auto [x, y] = q.front();
            q.pop();
            for (int k = 0; k < 4; k++) {
                int nx = x + dx[k], ny = y + dy[k];
                if (0 <= nx && nx < m && 0 <= ny && ny < n && board[nx][ny] == 'O') {
                    board[nx][ny] = 'E';
                    q.push({nx, ny});
                }
            }
        }
  
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                else if (board[i][j] == 'E') board[i][j] = 'O';
            }
        }
    }
  };