摘要：原題鏈接邊緣搜索替換法復雜度時間空間思路從矩陣的四條邊上的點作為起點，搜索連續的一塊區域上值為的點并賦為一個臨時變量。對四條邊上所有點都進行過這個步驟后，矩陣內剩余的就是被包圍的。

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

原題鏈接

時間 O(MN) 空間 O(MN)

從矩陣的四條邊上的點作為起點，搜索連續的一塊區域上值為O的點并賦為一個臨時變量。這里我們用BFS或DFS進行搜索。對四條邊上所有點都進行過這個步驟后，矩陣內剩余的O就是被包圍的O。此時再對所有點遍歷一遍，將臨時變量賦回O，而剩余的O則賦為X。

用隊列實現BFS時，賦臨時變量B時要在將周邊點加入隊列時就賦值，減少while循環的次數，否則Leetcode會超時

廣度優先 BFS

public class Solution {
    public void solve(char[][] board) {
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if((i == 0 || j == 0 || i == board.length - 1 || j == board[0].length - 1) && (board[i][j]=="O")){
                    Queue q = new LinkedList();
                    q.offer(i * board[0].length + j);
                    board[i][j] = "B";
                    while(!q.isEmpty()){
                        Integer key = q.poll();
                        int x = key / board[0].length;
                        int y = key % board[0].length;
                        if(x > 0 && board[x-1][y]=="O"){
                            q.offer((x-1) * board[0].length + y);
                            board[x-1][y] = "B";
                        }
                        if(x < board.length - 1  && board[x+1][y]=="O"){
                            q.offer((x+1) * board[0].length + y);
                            board[x+1][y] = "B";
                        }
                        if(y > 0  && board[x][y-1]=="O"){
                            q.offer(x * board[0].length + y - 1);
                            board[x][y-1] = "B";
                        }
                        if(y < board[0].length - 1  && board[x][y+1]=="O"){
                            q.offer(x * board[0].length + y + 1);
                            board[x][y+1] = "B";
                        }
                    }
                }
            }
        }
        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]=="B") board[i][j]="O";
            }
        }
    }
}

深度優先 DFS

public class Solution {
    static class Pair {
        public int first;
        public int second;
        public Pair(int f, int s) {
            first = f;
            second = s;
        }
    }
    
    public void solve(char[][] board) {
        if(board == null || board.length == 0) {
            return ;
        }
        for(int i = 0; i < board[0].length; ++i) {
            if(board[0][i] == "O") {
                markUnflippable(board,0,i);
            }
        }
        for(int i = 0; i < board[board.length-1].length; ++i) {
            if(board[board.length-1][i] == "O") {
                markUnflippable(board,board.length-1,i);
            }
        }
        for(int i = 0 ; i < board.length; ++i) {
            if(board[i][0] == "O") {
                markUnflippable(board,i,0);
            }
        }
        for(int i =0; i < board.length; ++i) {
            if(board[i][board[i].length-1] == "O") {
                markUnflippable(board,i,board[i].length-1);
            }
        }
    
        // modify the board
        for(int i = 0; i < board.length; ++i) {
            for(int j = 0; j < board[i].length; ++j) {
                if(board[i][j] == "O") {
                    board[i][j] = "X";
                } else if(board[i][j] == "U") {
                    board[i][j] = "O";
                }
            }
        }
    }
    
    public void markUnflippable(char[][] board, int r, int c) {
        int[] dirX = {-1,0,1,0};
        int[] dirY = {0,1,0,-1};
        ArrayDeque stack = new ArrayDeque<>();
        stack.push(new Pair(r,c));
        while(!stack.isEmpty()) {
            Pair p = stack.pop();
            board[p.first][p.second] = "U";
            for(int i = 0; i < dirX.length; ++i) {
                if(p.first + dirX[i] >= 0 && p.first + dirX[i] < board.length && p.second + dirY[i] >= 0 && p.second +dirY[i] < board[p.first + dirX[i]].length && board[p.first+dirX[i]][p.second+dirY[i]] == "O") {
                    stack.push(new Pair(p.first+dirX[i],p.second+dirY[i]));
                }
            }
        }
    }
}