[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 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

这个题目的思路分两步：

1. 把棋盘四周的"O"以及与其直接相连的“O”全部都变为“Q”，因为这种“O”是“X”围不住的。这里采用BFS算法。BFS我习惯的写法就是MIT算法课里面讲的，frontier加next进行while循环。要注意的是，把点放到frontier和next之前就要改变其状态，要不然会有重复的点被加进去。

2. 经过第一步之后，棋盘中还剩下的“O”就是可以被“X”围住的。所以接下来就要将“O”变为“X”，然后将“Q“变为”O“。这里要注意先后顺序，要不然第一步就白忙活了，又都变回去了。

代码如下：

public class Solution {    public void solve(char[][] board) {        if(board==null || board.length==0 || board[0].length==0) return;        flipOToQ(board);        flipOToX(board);        return;    }        public void flipOToQ(char[][] board) {        for(int i=0; i<board[0].length; i++) {            if(board[0][i] == 'O') {                flipOToQHelper(board, 0, i);            }            if(board[board.length-1][i] == 'O') {                flipOToQHelper(board, board.length-1, i);            }        }        for(int j=0; j<board.length; j++) {            if(board[j][0] == 'O') {                flipOToQHelper(board, j, 0);            }            if(board[j][board[0].length-1] == 'O') {                flipOToQHelper(board, j, board[0].length-1);            }        }        return;    }        // BFS, change state before adding into frontier and next    public void flipOToQHelper(char[][] board, int row, int col) {        Queue<Position> frontier = new LinkedList<Position>();        frontier.add(new Position(row, col));        board[row][col] = 'Q';        while(!frontier.isEmpty()) {            Queue<Position> next = new LinkedList<Position>();            while(!frontier.isEmpty()) {                Position temp = frontier.poll();                if(temp.col-1>=0 && board[temp.row][temp.col-1]=='O') {                    board[temp.row][temp.col-1] = 'Q';                    next.add(new Position(temp.row, temp.col-1));                }                if(temp.col+1<board[0].length && board[temp.row][temp.col+1]=='O') {                    board[temp.row][temp.col+1] = 'Q';                    next.add(new Position(temp.row, temp.col+1));                }                if(temp.row-1>=0 && board[temp.row-1][temp.col]=='O') {                    board[temp.row-1][temp.col] = 'Q';                    next.add(new Position(temp.row-1, temp.col));                }                if(temp.row+1<board.length && board[temp.row+1][temp.col]=='O') {                    board[temp.row+1][temp.col] = 'Q';                    next.add(new Position(temp.row+1, temp.col));                }            }            frontier = next;        }        return;    }        public void flipOToX(char[][] board) {        for(int i=0; i<board.length; i++) {            for(int j=0; j<board[0].length; j++) {                if(board[i][j]=='O') board[i][j]='X';                if(board[i][j]=='Q') board[i][j]='O';            }        }        return;    }}class Position {    public int row;    public int col;    public Position(int r, int c) {        this.row = r;        this.col = c;    }}