leetcode130

C++实现，使用BFS：

struct POS
{
    int x;
    int y;
    POS(int newx, int newy): x(newx), y(newy) {}
};

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        if(board.empty() || board[].empty())
            return;
        int m = board.size();
        int n = board[].size();
        for(int i = ; i < m; i ++)
        {
            for(int j = ; j < n; j ++)
            {
                if(board[i][j] == 'O')
                {
                    if(i ==  || i == m- || j ==  || j == n-)
                    {// remain 'O' on the boundry
                        bfs(board, i, j, m, n);
                    }
                }
            }
        }
        for(int i = ; i < m; i ++)
        {
            for(int j = ; j < n; j ++)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == '*')
                    board[i][j] = 'O';
            }
        }
    }
    void bfs(vector<vector<char>> &board, int i, int j, int m, int n)
    {
        stack<POS*> stk;
        POS* pos = new POS(i, j);
        stk.push(pos);
        board[i][j] = '*';
        while(!stk.empty())
        {
            POS* top = stk.top();
            if(top->x >  && board[top->x-][top->y] == 'O')
            {
                POS* up = new POS(top->x-, top->y);
                stk.push(up);
                board[up->x][up->y] = '*';
                continue;
            }
            if(top->x < m- && board[top->x+][top->y] == 'O')
            {
                POS* down = new POS(top->x+, top->y);
                stk.push(down);
                board[down->x][down->y] = '*';
                continue;
            }
            if(top->y >  && board[top->x][top->y-] == 'O')
            {
                POS* left = new POS(top->x, top->y-);
                stk.push(left);
                board[left->x][left->y] = '*';
                continue;
            }
            if(top->y < n- && board[top->x][top->y+] == 'O')
            {
                POS* right = new POS(top->x, top->y+);
                stk.push(right);
                board[right->x][right->y] = '*';
                continue;
            }
            stk.pop();
        }
    }
};

补充一个python的实现，使用DFS：

 class Solution:
     def dfs(self,board,i,j,m,n,visited,direction):
         if i < 0 or i >= m or j < 0 or j >= n or visited[i][j] == 1:
             return
         if board[i][j] == 'O':
             board[i][j] = '*'
             visited[i][j] = 1

             for dirt in direction:
                 if dirt[0] == 0 and dirt[1] == 0:
                     continue
                 x = i + dirt[0]
                 y = j + dirt[1]
                 self.dfs(board,x,y,m,n,visited,direction)

     def solve(self, board: 'List[List[str]]') -> None:
         m = len(board)#行
         if m == 0:
             return
         n = len(board[0])#列
         if n == 0:
             return
         visited = [[0 for _ in range(n)]for _ in range(m)]
         direction = [[-1,0],[1,0],[0,-1],[0,1]]
         for i in range(m):
             for j in range(n):
                 if board[i][j] == 'O' and (i == 0 or i == m-1 or j == 0 or j == n-1):
                     self.dfs(board,i,j,m,n,visited,direction)

         for i in range(m):
             for j in range(n):
                 if board[i][j] == 'O':
                     board[i][j] = 'X'
                 elif board[i][j] == '*':
                     board[i][j] = 'O'