LeetCode: Surrounded Regions

思路：刚开始觉得如果每个'O'点如果四个方向上存在'X'点，则其被包围，其实如果存在一条路径，即所有'O'点相连，最后连接到边界上去了，那么这个点就没有被包围，所以，只能通过边界的'O'点来进行深度搜索，只要与边界'O'相连通的'O'点都是没有被覆盖的，剩下的就好办了。

code:

class Solution {public:    bool isRightLoc(int i,int j,int rows,int cols){        if(i <rows && i >=0 && j < cols && j >=0)            return true;        return false;    }    void solve(vector<vector<char>> &board) {        if(board.size() == 0)return;        int rows = board.size(), cols = board[0].size();        for(int j = 0;j < cols;j++){            if(board[0][j] == 'O'){               //board[0][j] = 'T';//cout<<"sdf"<<endl;                dfs(0,j,board);            }            if(board[rows-1][j] == 'O'){                //board[rows-1][j] = 'T';                dfs(rows-1,j,board);            }        }        for(int i = 1;i < rows-1;i++){            if(board[i][0] == 'O'){                //board[i][0] = 'T';                dfs(i,0,board);            }            if(board[i][cols-1] == 'O'){                //board[i][cols-1] = 'T';                dfs(i,cols-1,board);            }        }                for(int i = 0;i < rows;i++)            for(int j = 0;j < cols;j++){                if(board[i][j] == 'O')                    board[i][j] = 'X';                else if(board[i][j] == 'T')                    board[i][j] = 'O';            }    }    void dfs(int i,int j,vector<vector<char>> &board){        int rows = board.size(), cols = board[0].size();        if(isRightLoc(i,j,rows,cols)){//cout<<i<<" "<<j<<" "<<board[i][j]<<endl;            if(board[i][j] == 'O'){                board[i][j] = 'T';                dfs(i,j+1,board);                dfs(i-1,j,board);                dfs(i,j-1,board);                dfs(i+1,j,board);            }        }    }};