Surrounded Regions

来源：互联网发布：淘宝客服培训流程编辑：程序博客网时间：2024/05/17 22:07

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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

用BFS，记录每一个和边界O相连的O。

class Solution {public:    void solve(vector<vector<char> > &board) {        int width = board.size();        if (width == 0)            return;        int length = board[0].size();        if  (length == 0)            return;            for (int i = 0; i < length; ++i)        {            if (board[0][i] == 'O')                bfsBoundary(board, 0, i);                if (board[width-1][i] == 'O')                bfsBoundary(board, width-1, i);        }            for (int i = 0; i < width; ++i)        {            if (board[i][0] == 'O')                bfsBoundary(board, i, 0);            if (board[i][length-1] == 'O')                bfsBoundary(board, i, length-1);        }            for (int i = 0; i < width; ++i)        {            for (int j = 0; j < length; ++j)            {                if (board[i][j] == 'O')                    board[i][j] = 'X';                else if (board[i][j] == 'B')                    board[i][j] = 'O';            }        }    }        void bfsBoundary(vector<vector<char> >& board, int w, int l)    {        int width = board.size();        int length = board[0].size();        queue<pair<int, int> > q;        q.push(pair<int, int>(w, l));        board[w][l] = 'B';        while (!q.empty()) {            pair<int, int> cur = q.front();            q.pop();            int adjs[4][2] = {{-1, 0},                 {1, 0},                 {0, -1},                {0, 1}};            for (int i = 0; i < 4; ++i)            {                int adjW = adjs[i][0] + cur.first;                int adjL = adjs[i][1] + cur.second;                if ((adjW >= 0) && (adjW < width) && (adjL >= 0)                        && (adjL < length)                         && (board[adjW][adjL] == 'O')) {                    q.push(make_pair(adjW, adjL));                    board[adjW][adjL] = 'B';                }            }        }    }};

0 0