130. Surrounded Regions

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这道题只要把与边缘相连的‘O’全部换成‘V’，然后将所有的除‘V’之外的字母全部换成‘X’,最后再讲所有的‘V’换成‘O’。方法的就是将DFS与边缘相连的‘O’, 或者BFS与边缘相连的‘O’。但是递归版本的DFS栈溢出，所有用栈的DFS即可。
1. 超时版本

class Solution {public:void solve(vector<vector<char>>& board) {    if(board.empty() || board[0].empty()) return;    int n = board.size(), m = board[0].size();    for(int i = 0; i < n; i++)        for(int j = 0; j < m; j++)            if(i == 0 || i == n-1 || j == 0 || j == m-1)                dfs(board, i, j);    for(int i = 0; i < n; i++)        for(int j = 0; j < m; j++)            board[i][j] = (board[i][j] == 'V') ? 'O' : 'X';}void dfs(vector<vector<char>> &board, int row, int col){    int n = board.size(), m = board[0].size();    if(row < 0 || row >= n || col < 0 || col >= m) return;    if(board[row][col] != 'O') return;    board[row][col] = 'V';    dfs(board, row+1, col);    dfs(board, row-1, col);    dfs(board, row, col+1);    dfs(board, row, col-1);    }};

改进版本用栈

 class Solution {    public:        typedef vector<vector<char> > BOARDTYPE;        void solve(BOARDTYPE &board) {            if (board.empty() || board[0].empty()) return;            int N = board.size(), M = board[0].size();            for (int i = 0; i < N; ++i)                for (int j = 0; j < M; ++j)                    if (i == 0 || j == 0 || i == N-1 || j == M-1)                        dfs(board, i, j); // you may call dfs or bfs here!            for (int i = 0; i < N; ++i)                for (int j = 0; j < M; ++j)                    board[i][j] = (board[i][j] == 'V') ? 'O' : 'X';        }        void dfs(vector<vector<char>> &board, int row, int col)        {            int n = board.size(), m = board[0].size();            if(row < 0 || row >= n || col < 0 || col >= m) return;            if(board[row][col] != 'O') return;            stack<pair<int, int>> s;            s.push(make_pair(row, col));            while(!s.empty())            {                row = s.top().first;                col = s.top().second;                s.pop();                board[row][col] = 'V';                if(row+1 < n && board[row+1][col] == 'O')                    s.push(make_pair(row+1, col));                if(row-1 >= 0 && board[row-1][col] == 'O')                    s.push(make_pair(row-1, col));                if(col+1 < m && board[row][col+1] == 'O')                    s.push(make_pair(row, col+1));                if(col-1 >= 0 && board[row][col-1] == 'O')                    s.push(make_pair(row, col-1));            }        }    };

BFS方法

 class Solution {    public:        typedef vector<vector<char> > BOARDTYPE;        void solve(BOARDTYPE &board) {            if (board.empty() || board[0].empty()) return;            int N = board.size(), M = board[0].size();            queue<pair<int,int>> q;            for (int i = 0; i < N; ++i){                if(board[i][M-1]=='O' ){                    q.push(make_pair(i,M-1));                    board[i][M-1]='V';                     }                    if(board[i][0]=='O'){                    q.push(make_pair(i,0));                    board[i][0]='V';                }            }            for (int j = 0; j < M; ++j){                if(board[0][j]=='O'){                    q.push(make_pair(0,j));                        board[0][j]='V';                }                     if( board[N-1][j]=='O'){                    q.push(make_pair(N-1,j));                    board[N-1][j]='V';                }            }            bfs(board,q);            for (int i = 0; i < N; ++i)                for (int j = 0; j < M; ++j)                    board[i][j] = (board[i][j] == 'V') ? 'O' : 'X';        }         void bfs(BOARDTYPE &board, queue<pair<int,int>> &q) {            int N = board.size(), M = board[0].size();            while(!q.empty()){                int row = q.front().first;                int col = q.front().second;                q.pop();                if(row+1<N && board[row+1][col]=='O'){                    q.push(make_pair(row+1,col));                    board[row+1][col]='V';                }                if(row-1>=0 && board[row-1][col]=='O'){                    q.push(make_pair(row-1,col));                    board[row-1][col]='V';                }                if(col+1<M && board[row][col+1]=='O'){                    q.push(make_pair(row,col+1));                    board[row][col+1]='V';                }                if(col-1<N && board[row][col-1]=='O'){                    q.push(make_pair(row,col-1));                    board[row][col-1]='V';                }            }//end while        }    };

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