【LeetCode】Surrounded Regions

Surrounded Regions 
Total Accepted: 4485 Total Submissions: 31870 My Submissions
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
Discuss
        个人感觉这个题目和九度题目1460：Oil Deposit很像，解题思路参考(Jobdu 题目1460：Oil Deposit)，各位不妨思考一下相似之处。
        好吧，我承认第一遍看这个题目有点不太懂。
        其实题目大意是这样的，从一个二维矩阵中寻找所有被X包围的O，如果找到的话，将所有的O替换为X。
        BFS，基本思路是从边界开始找起，同时标记是否访问过，防止重复搜索。如果边界有O的话，从该点开始宽度优先搜索，搜索到O就标记访问过。针对每个边界上为O的点搜索，这样就可以搜索到结果。搜索完毕之后，所有的为O的未访问过的点，都为被X包围的点，直接赋值为X即可。

Java AC

public class Solution {    public int stepArr[][] = {{-1,0},{1,0},{0,-1},{0,1}};    public int visit[][];    public int m, n;    public void solve(char[][] board) {        if(board == null || board.length == 0){            return;        }        m = board.length;        n = board[0].length;        visit = new int[m][n];        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if((i == 0 || i == m-1 || j == 0 || j == n-1)                             && visit[i][j] == 0 && board[i][j] == 'O'){                    bfs(i,j,board);                }            }        }        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(visit[i][j] != 1 && board[i][j] == 'O'){                    board[i][j] = 'X';                }            }        }    }    public void bfs(int nodei, int nodej, char board[][]){        Queue<Node> queue = new LinkedList<Node>();        Node node = new Node(nodei, nodej);        queue.offer(node);        visit[nodei][nodej] = 1;        while(!queue.isEmpty()){            node = queue.peek();            queue.poll();            int newx = 0;            int newy = 0;            for(int i = 0; i < 4; i++){                newx = node.x + stepArr[i][0];                newy = node.y + stepArr[i][1];                if(newx >= 0 && newx < m && newy >= 0 && newy < n                             && visit[newx][newy] == 0 && board[newx][newy] == 'O'){                    visit[newx][newy] = 1;                    queue.offer(new Node(newx, newy));                }            }        }    }    public class Node{        int x;        int y;        Node(int x, int y){            this.x = x;            this.y = y;        }    }}