【LeetCode】130. Surrounded Regions

最新推荐文章于 2023-12-01 14:27:39 发布

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本文探讨了一个二维棋盘中'O'被'X'围困的问题，提出了一种利用深度优先搜索（DFS）的解决方案。通过从棋盘边界上的'O'开始进行DFS，标记不会被围困的'O'，最后将所有未标记的'O'转换为'X'，实现了对被围困区域的有效捕获。

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.

题解：找出被包围的o并转为x，本题用dfs很简单，但是需要注意入手角度，刚开始只从找出被包围的o入手结果要考虑边界的o以及形成o的并查集比较麻烦，所以应该先从边界四条的o开始dfs，并把这些o标记为1，然后遍历图，把所有o改为x，把1改为o即可

代码如下：

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        
        int n=board.size();
        if(n==0||n==1||n==2) return;
        int m=board[0].size();
        for(int i=0;i<n;i++){
            dfs(i,0,board,n,m);
            dfs(i,m-1,board,n,m);
        }
        for(int j=0;j<m;j++){
            dfs(0,j,board,n,m);
            dfs(n-1,j,board,n,m);
        }
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++){
                if(board[i][j]=='O') board[i][j]='X';
                if(board[i][j]=='1') board[i][j]='O';
            }
        
    }
    void dfs(int i,int j,vector<vector<char>>& board,int n,int m){
        if(i<0||j<0||i>=n||j>=m) return;
        if(board[i][j]=='O'){
            board[i][j]='1';
            dfs(i+1,j,board,n,m);
            dfs(i-1,j,board,n,m);
            dfs(i,j+1,board,n,m);
            dfs(i,j-1,board,n,m);
        }
         
    }
};