[Lintcode]Surrounded Regions

最新推荐文章于 2020-05-03 05:58:04 发布

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本文介绍了一种解决二维棋盘中被X包围的O型区域捕获问题的算法。通过先标记边界上的O及其相连的O，再将剩余的O转化为X，最后恢复边界O为O的方法实现。此算法简化了问题处理过程。

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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.

Example

X X X X
X O O X
X X O X
X O X X

After capture all regions surrounded by 'X', the board should be:

X X X X
X X X X
X X X X
X O X X

分析：首先标记所有不被包围的O，比如改为F。对每一个不被包围的O，与其相连的点肯定不被包围。然后遍历，将剩余的O改为X。之后将F恢复为O。如果正向考虑，每一个点判断其是否被包围，那么需要考虑的情况非常多，需要考虑每一个相邻的点是否被包围，从而需要确定所有相邻的点不会遇到边界。但是反过来标记不被包围的点，则简单一些

public class Solution {
    /**
     * @param board a 2D board containing 'X' and 'O'
     * @return void
     */
    public void surroundedRegions(char[][] board) {
        if(board.length == 0) return;
        if(board[0].length == 0) return;
        
        for(int i = 0; i < board.length; i++) {
            if(board[i][0] == 'O') {
                fill(board, i, 0);
            }
            if(board[i][board[0].length - 1] == 'O') {
                fill(board, i, board[0].length - 1);
            }
        }
        
        for(int i = 0; i < board[0].length; i++) {
            if(board[0][i] == 'O') {
                fill(board, 0, i);
            }
            if(board[board.length - 1][i] == 'O') {
                fill(board, board.length - 1, i);
            }   
        }
        
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                if(board[i][j] == 'O') board[i][j] = 'X';
                if(board[i][j] == 'F') board[i][j] = 'O';
            }
        }
    }
    
    void fill(char[][] board, int x, int y) {
        if(x < 0 || x >= board.length || y < 0 || y >= board[0].length) return;
        if(board[x][y] != 'O') return;
        
        board[x][y] = 'F';
        fill(board, x + 1, y);
        fill(board, x - 1, y);
        fill(board, x, y - 1);
        fill(board, x, y + 1);
    }
}