LEETCODE: Surrounded Regions-优快云博客

本文介绍了一种解决二维棋盘中被'X'包围的所有'O'区域问题的算法实现。通过从棋盘边界开始遍历，并使用广度优先搜索（BFS）标记与边界相连的'O'，最终将未被标记的'O'转变为'X'，完成围捕。

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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.

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

这题对速度还是有要求的，所以只能从四条边往里面找。想用偷懒逐行扫描是不行的！

class Solution {
public:
    int dimy;
    int dimx;
    void travelling(int dy, int dx, vector<vector<char>> &board) {
        if(board[dy][dx] == 'X') return;
        queue<pair<int, int>> thequeue;
        thequeue.push(make_pair(dy, dx));
        while(!thequeue.empty()) {
            pair<int, int> front = thequeue.front();
            thequeue.pop();
            board[front.first][front.second] = 'Z';
            if(front.first > 0 && board[front.first - 1][front.second] == 'O') {
                board[front.first - 1][front.second] = 'Z';
                thequeue.push(make_pair(front.first - 1, front.second));
            }
            if(front.first < dimy - 1 && board[front.first + 1][front.second] == 'O') {
                board[front.first][front.second] = 'Z';
                thequeue.push(make_pair(front.first + 1, front.second));
            }
            if(front.second > 0 && board[front.first][front.second - 1] == 'O') {
                board[front.first][front.second - 1] = 'Z';
                thequeue.push(make_pair(front.first, front.second - 1));  
            }
            if(front.second < dimx - 1 && board[front.first][front.second + 1] == 'O') {
                board[front.first][front.second + 1] = 'Z';
                thequeue.push(make_pair(front.first, front.second + 1));  
            }
        }
    }
    void solve(vector<vector<char>> &board) {
        dimy = board.size();
        if(dimy == 0) return;
        dimx = board[0].size();
        if(dimx == 0) return;
        for(int ii = 0; ii < dimy; ii ++) {
            travelling(ii, 0, board);
            travelling(ii, dimx - 1, board);
        }
        for(int ii = 0; ii < dimx; ii ++) {
            travelling(0, ii, board);
            travelling(dimy - 1, ii, board);
        }
        
        for(int ii = 0; ii < dimy; ii ++) {
            for(int jj = 0; jj < dimx; jj ++) {
                board[ii][jj] = board[ii][jj] == 'Z' ? 'O' : 'X';
            }
        }
    }
};