leetcoe || 130、Surrounded Regions

problem：

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

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题意：将被X包围的O换成X

thinking：

（1）首先我用递归做的，提交栈溢出了....

（2）参考网上的实现，要用到一种图形学的算法：泛洪算法-维基

采用BFS或者DFS，使用queue或者stack结构。

code：

class Solution {public:    int row;    int col;    void bfs(vector<vector<char> > &board, int r, int c) {        queue<pair<int, int> > qe;        qe.push({r, c});        while(!qe.empty()) {                    r = qe.front().first;            c = qe.front().second;            qe.pop();            if (r>0&&board[r-1][c]=='O') {                board[r-1][c] = '+';                qe.push({r-1,c});            }            if (r+1<row&&board[r+1][c]=='O') {                board[r+1][c] = '+';                qe.push({r+1,c});            }            if (c>0&&board[r][c-1]=='O') {                board[r][c-1] = '+';                qe.push({r,c-1});            }            if (c+1<col&&board[r][c+1]=='O') {                board[r][c+1] = '+';                qe.push({r,c+1});            }        }        return;    }    void solve(vector<vector<char> > &board) {        if (board.empty())            return;        row = board.size();                col = board[0].size();                // top edge        for(int i = 0; i < col; ) {            if(board[0][i] == 'O') {                int j=i;                while(board[0][j]=='O') j++;                board[0][i] = '+';                bfs(board, 0, i);                i = j;            } else                 i++;        }        // bottom edge        for(int i = 0; i < col;) {            if(board[row-1][i] == 'O') {                int j=i;                while(board[row-1][j]=='O') j++;                board[row-1][i] = '+';                bfs(board, row-1, i);                i = j;            } else                i++;        }        // left edge        for(int i = 1; i < row-1;) {            if(board[i][0] == 'O') {                int j=i;                while(board[j][0]=='O') j++;                board[i][0] = '+';                bfs(board, i, 0);                i = j;            } else                i++;        }        // right edge        for(int i = 1; i < row-1; ) {            if(board[i][col-1] == 'O') {                int j=i;                while(board[j][col-1]=='O') j++;                board[i][col-1] = '+';                bfs(board, i, col-1);                i = j;            } else                i++;        }        // then scan all the cells, recover live cells and flip dead cells         for(int i = 0; i < row; i++)             for(int j = 0; j < col; j++)                 board[i][j] = (board[i][j]=='+')?'O':'X';        return;    }};