[leetcode]Surrounded Regions

题目描述：

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

题目分析：

首先找到从四个边界开始寻找O,将其进行标记。标记完后，对于剩余的的O就将其变为X，标记的进行还原即可。

代码：

class Solution {public:    void solve(vector<vector<char>> &board) {        if(board.size()==0)return;        int row=board.size();        int col=board[0].size();        //up        for(int i=0;i<col;i++){            if(board[0][i]=='O'){                bfs(board,0,i);            }        }        //right        for(int i=0;i<row;i++){            if(board[i][col-1]=='O'){                bfs(board,i,col-1);            }        }        //down        for(int i=0;i<col;i++){            if(board[row-1][i]=='O'){                bfs(board,row-1,i);            }        }        //left        for(int i=0;i<row;i++){            if(board[i][0]=='O'){                bfs(board,i,0);            }        }        //modify        for(int i=0;i<row;i++){            for(int j=0;j<col;j++){                if(board[i][j]=='+'){                    board[i][j]='O';                } else if(board[i][j]=='O'){                    board[i][j]='X';                }            }        }             }    //search bounds    void bfs(vector<vector<char>> &board, int i, int j) {        queue<int> q;        int col=board[0].size();        visit(board,i,j,q);        while(!q.empty()){            int  cur=q.front();            q.pop();            int x=cur/col;            int y=cur%col;            visit(board,x-1,y,q);            visit(board,x,y-1,q);            visit(board,x+1,y,q);            visit(board,x,y+1,q);        }}    void visit(vector<vector<char>> &board, int i, int j, queue<int> &q) {        int m = board.size();        int n = board[0].size();        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != 'O')            return;        board[i][j] = '+';        q.push(i * n + j);    }};