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  • 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

    [解题思路]

     board的四个边上的O是无法被X包围的,故如果在四个边上发现O,则这些O应当被保留 从这些O出发继续寻找相邻的O,这些O也是要保留的

     等这些O都标记结束,则剩余的O就应该被改成X

    public class Solution {
        public void solve(char[][] board) {
          
            if(board == null||board.length == 0||board[0] == null||board[0].length == 0)
                return;
            int y = board.length;
            int x = board[0].length;
            LinkedList<location> mylist = new LinkedList<location>();
            for(int i = 0; i < x; i++)
            {
                // 第一行
                if(board[0][i] == 'O')
                {
                    board[0][i] = 'Y';
                    mylist.add(new location(0, i));
                }
                // 最后一行
                if(board[y-1][i] == 'O')
                {
                    board[y-1][i] = 'Y';
                    mylist.add(new location(y-1, i));
                }
            }
            for(int i = 0; i < y; i++)
            {   
                // 第一列
                if(board[i][0] == 'O')
                {
                    board[i][0] = 'Y';
                    mylist.add(new location(i, 0));
                }
                //最后一列
                if(board[i][x-1] == 'O')
                {
                    board[i][x-1] = 'Y';
                    mylist.add(new location(i, x-1));
                }
            }
            while(!mylist.isEmpty())
            {
                location myloc = mylist.poll();
                if(myloc.x - 1 >= 0 && board[myloc.x - 1][myloc.y] == 'O')
                {
                    board[myloc.x - 1][myloc.y] = 'Y';
                    mylist.add(new location(myloc.x - 1, myloc.y));
                }
                if(myloc.y - 1 >= 0 && board[myloc.x][myloc.y - 1] == 'O')
                {
                    board[myloc.x][myloc.y - 1] = 'Y';
                    mylist.add(new location(myloc.x, myloc.y - 1));
                }
                if(myloc.x + 1 < y && board[myloc.x + 1][myloc.y] == 'O')
                {
                    board[myloc.x + 1][myloc.y] = 'Y';
                    mylist.add(new location(myloc.x + 1, myloc.y));
                }
                if(myloc.y + 1 < x && board[myloc.x][myloc.y + 1] == 'O')
                {
                    board[myloc.x][myloc.y + 1] = 'Y';
                    mylist.add(new location(myloc.x, myloc.y + 1));
                }
            }
            
            for(int i = 0; i < y; i++){
                for(int j = 0; j < x; j++)
                {
                    if(board[i][j] == 'Y')
                        board[i][j] = 'O';
                    else if(board[i][j] == 'O')
                        board[i][j] = 'X';
                }    
            }
        }
        class location{
            int x;
            int y;
            location(int x, int y)
            {this.x = x; this.y = y;}
        }
    }
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  • 原文地址:https://www.cnblogs.com/RazerLu/p/3537750.html
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