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  • leetcode--Surrounded Regions

    Given a 2D board containing 'X' and '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

    This is a simple problem, we just need to apply bfs on 'O' elements on the edges of board.
    public class Solution {
        public void solve(char[][] board) {
            int row = board.length;
    		if(row > 0){
    			int column = board[0].length;
    			Set<Integer> alreadyChecked = new HashSet<Integer>();
    			for(int i = 0; i < column; ++i){
    				if(board[0][i] == 'O' && !alreadyChecked.contains(i))
    					bfs(board,0, i, alreadyChecked);
    				if(board[row - 1][i] == 'O' && !alreadyChecked.contains((row - 1) * column + i))
    					bfs(board,row - 1, i, alreadyChecked);
    			}
    			for(int i = 1; i < row - 1; ++i){
    				if(board[i][0] == 'O' && !alreadyChecked.contains(i * column))
    					bfs(board,i, 0, alreadyChecked);
    				if(board[i][column - 1] == 'O' && !alreadyChecked.contains(i * column + column - 1))
    					bfs(board,i, column - 1, alreadyChecked);
    			}
    			
    			for(int i = 0; i < row; ++i){
    				for(int j = 0; j < column; ++j){
    					if(board[i][j] == 'O' && !alreadyChecked.contains(i*column + j))
    						board[i][j] = 'X';
    				}
    			}
    		}
        }
    	
    	private void bfs(char[][] board, int i, int j, Set<Integer> alreadyChecked) {
    		int row = board.length;
    		if(row > 0){
    			int column = board[0].length;
    			if(!alreadyChecked.contains(i * column + j)) {
    				Queue<Integer> current = new LinkedList<Integer>();
    				alreadyChecked.add(i * column + j);
    				current.add(i * column + j);
    				while(current.peek() != null){
    					int currentCoordinate = current.poll();
    					int x = currentCoordinate / column;
    					int y = currentCoordinate % column;
    					
    					if(x - 1 >= 0 && board[x - 1][y] == 'O' && !alreadyChecked.contains(currentCoordinate - column)){
    						current.add(currentCoordinate - column);
    						alreadyChecked.add(currentCoordinate - column);
    					}
    					
    					if(x + 1 < row && board[x + 1][y] == 'O' && !alreadyChecked.contains(currentCoordinate + column)){
    						current.add(currentCoordinate + column);
    						alreadyChecked.add(currentCoordinate + column);
    					}
    					
    					if(y - 1 >= 0 && board[x][y - 1] == 'O' && !alreadyChecked.contains(currentCoordinate - 1)) {
    						current.add(currentCoordinate - 1);
    						alreadyChecked.add(currentCoordinate - 1);
    					}
    					
    					if(y + 1 < column && board[x][y + 1] == 'O' && !alreadyChecked.contains(currentCoordinate + 1)) {
    						current.add(currentCoordinate + 1);
    						alreadyChecked.add(currentCoordinate + 1);
    					}
    				}
    			}
    		}    
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/averillzheng/p/3825371.html
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