LeetCode_Surrounded Regions

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 X
X O O X
X X O X
X O X X
After running your function, the board should be: X X X X
X X X X
X X X X
X O X X

  solution: 思想是mark 出所有外围的O,里面的不管。

  1. First scan the four edges of the board, if you meet an 'O', call a recursive mark function to mark that region to something else (for example, '+');
  2. scan all the board, if you meet an 'O', flip it to 'X';
  3. scan again the board, if you meet an '+', flip it to 'O';
    class Solution {
    public:
    void BFS(vector<vector<char>> &board, int i , int j){ if(board[i][j] == 'X' || board[i][j] == '#') return;
    board[i][j] = '#'; if( i - >= ) BFS(board, i-, j );
    if( i + < rows ) BFS(board, i+, j );
    if( j - >= ) BFS(board, i, j- );
    if( j + < columns ) BFS(board, i, j+ );
    }
    void solve(vector<vector<char>> &board) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    rows = board.size();
    if( rows == ) return ;
    columns = board[].size();
    if(columns == ) return ; for( int i = ; i < columns - ; i++)
    {
    BFS(board, ,i);//up
    BFS(board, rows - , i );//down
    }
    for( int i = ; i< rows; i++)
    {
    BFS(board, i, ); //left
    BFS(board, i, columns - ); //right
    }
    for(int i = ; i< rows; i++)
    for(int j = ; j < columns ; j++)
    {
    if(board[i][j] == '#')
    board[i][j] = 'O';
    else if (board[i][j] == 'O')
    board[i][j] = 'X';
    }
    }
    private :
    int rows;
    int columns;
    };
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