130. Surrounded Regions (DFS)

最新推荐文章于 2021-01-22 06:58:09 发布

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本文介绍了一个LeetCode经典题目“围攻区域”的深度优先搜索(DFS)解决方案。该题要求将被 'X' 包围的 'O' 转换为 'X'，但边界上的 'O' 或与之相连的 'O' 除外。通过从边界开始的DFS遍历，标记并最终转换矩阵中的字符，实现了这一目标。

经过一个月的倦怠期，我徐汉三又回来啦！FLAG每天至少一道题坚持到秋招offer！！！

今天的题目是DFS思想。https://leetcode.com/problems/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.

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

Explanation:

Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

意思就是边界上的‘O’以及和边界上的‘O’相邻的‘O’ 可以保留，其余的转成‘X’.

所以可以从四个边界出发进行深度优先遍历，标记和边界上‘O’相邻的“O”为‘*’；

然后依次遍历所有点，将“*”变成‘O’,其他的都是‘X’.

注意不要忘记边界条件判断。

class Solution {
    private int[][] directions = {{0,1},{0,-1},{1,0},{-1,0}};
    
    public void solve(char[][] board) {
        if(board==null || board.length==0)
            return;
        for(int i=0;i<board.length;i++){
            dfs(board, i, 0);
            dfs(board, i, board[0].length-1);
        }
        for(int i=0;i<board[0].length;i++){
            dfs(board, 0, i);
            dfs(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]=='*')
                    board[i][j]='O';
                else
                    board[i][j]='X';
            }
        }
    }
    public void dfs(char[][] board, int i, int j){
        if(i<0 || i>= board.length || j<0 || j>=board[0].length || board[i][j]!='O')
            return;
        board[i][j]='*';
        for(int[] dir:directions){
            dfs(board, i+dir[0], j+dir[1]);
        }
    }
}