[LeetCode] Surrounded Regions

最新推荐文章于 2020-08-11 10:37:01 发布

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本文详细介绍了如何利用BFS算法解决二维板上被X包围的O区域翻转问题，避免了DFS算法可能导致的超时情况。通过代码实现和实例演示，阐述了不同算法的效率对比。

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

用bfs没问题

class Solution {
public:
    int m,n;
    queue<int> q;
    void add(int x,int y,vector<vector<char>> &board){
        if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O'){
            board[x][y] = 'q';
            q.push(x*n + y);
        }
    }
    void bfs(int x,int y,vector<vector<char>> &board){
        add(x,y,board);
        while(!q.empty()){
            int p = q.front();
            q.pop();
            int px = p / n,py = p % n;
            add(px - 1,py,board);
            add(px + 1,py,board);
            add(px,py - 1,board);
            add(px,py + 1,board);
        }
    }
    void solve(vector<vector<char>> &board) {
        if(board.empty()) return;
        m = board.size();
        n = board[0].size();
        for(int i = 0;i < n;i ++){
            bfs(0,i,board);
            bfs(m - 1,i,board);
        }
        for(int i = 0;i < m;i ++){
            bfs(i,0,board);
            bfs(i,n - 1,board);
        }
        for(int i = 0;i < m;i ++){
            for(int j = 0;j < n;j ++){
                board[i][j] = (board[i][j] == 'q' ? 'O' : 'X');
            }
        }
    }
};

用dfs会超时

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