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  • 130. Surrounded Regions

    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.

    For 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

    DFS的时候用两个list存访问过的地址,如果没有touch过边界,则用这两个list的值来给board改写。但是这个耗费的space比较大,过不了OJ的大的test case。

    public void Solve(char[,] board) {
            int row = board.GetLength(0);
            int col = board.GetLength(1);
            bool[,] visited = new bool[row,col];
            
            for(int i =0;i< row;i++)
            {
                for(int j =0;j<col;j++)
                {
                    if(board[i,j] == 'O' && !visited[i,j])
                    {
                        List<int> rowStore = new List<int>();
                        List<int> colStore = new List<int>();
                        bool NotWrap= BackTracking(board,i,j,row,col,visited,rowStore,colStore);
                        if(!NotWrap)
                        {
                            for(int m = 0;m< rowStore.Count();m++)
                            {
                                board[rowStore[m],colStore[m]] ='X';
                            }
                        }
                    }
                }
            }
            return;
        }
        
        private bool BackTracking(char[,] board, int x, int y, int row, int col, bool[,] visited, List<int> rowStore, List<int> colStore)
        {
            if(x<0 || x> row-1) return true;
            if(y<0 || y> col-1) return true;
            if(visited[x,y]) return false;
            if(x == 0 || x == row-1)
            {
                visited[x,y] = true;
                if(board[x,y] == 'O') 
                    return true;
            }
            if(y == 0 || y == col-1)
            {
                visited[x,y] = true;
                if(board[x,y] == 'O') return true;
            }
          
            if(board[x,y] == 'X') return false;
            visited[x,y] = true; 
            rowStore.Add(x);
            colStore.Add(y);
            return BackTracking(board,x+1,y,row,col,visited,rowStore,colStore) || BackTracking(board,x-1,y,row,col,visited,rowStore,colStore) || BackTracking(board,x,y+1,row,col,visited,rowStore,colStore) || BackTracking(board,x,y-1,row,col,visited,rowStore,colStore);
        }
        
        

    用边界点为‘O’的开始做DFS, 找到的做标记‘#’。然后第二次遍历,标记为‘#’的全换位‘O’, 标记为‘O’的换位‘X’.

     public void Solve(char[,] board) {
            int row = board.GetLength(0);
            int col = board.GetLength(1);
            
            for(int i =0;i< row;i++)
            {
                for(int j =0;j<col;j++)
                {
                    if(((i== 0 || i== row-1) ||(j == 0|| j ==col-1))&&(board[i,j] == 'O'))
                    {
                        BackTracking(board,i,j,row,col);
                    }
                }
            }
            
            for(int i =0;i< row;i++)
            {
                for(int j =0;j<col;j++)
                {
                    if(board[i,j] == 'O') board[i,j] = 'X';
                    else if(board[i,j] == '#') board[i,j] = 'O';
                }
            }
            return;
        }
        
        private void BackTracking(char[,] board, int x, int y, int row, int col)
        {
    
            //if(board[x,y] != 'O') return ;
            
            board[x,y] ='#';
            
            if(x<row-1 && board[x+1,y]=='O') BackTracking(board,x+1,y,row,col);
            if(x>0 && board[x-1,y]=='O') BackTracking(board,x-1,y,row,col);
            if(y<col-1 && board[x,y+1]=='O') BackTracking(board,x,y+1,row,col);
            if(y>0 && board[x,y-1]=='O') BackTracking(board,x,y-1,row,col);
        }
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  • 原文地址:https://www.cnblogs.com/renyualbert/p/5863447.html
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