[Leetcode] Surrounded Regions 找出被包围的区域

摘要：原题链接边缘搜索替换法复杂度时间空间思路从矩阵的四条边上的点作为起点，搜索连续的一块区域上值为的点并赋为一个临时变量。对四条边上所有点都进行过这个步骤后，矩阵内剩余的就是被包围的。

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

原题链接

时间 O(MN) 空间 O(MN)

从矩阵的四条边上的点作为起点，搜索连续的一块区域上值为O的点并赋为一个临时变量。这里我们用BFS或DFS进行搜索。对四条边上所有点都进行过这个步骤后，矩阵内剩余的O就是被包围的O。此时再对所有点遍历一遍，将临时变量赋回O，而剩余的O则赋为X。

用队列实现BFS时，赋临时变量B时要在将周边点加入队列时就赋值，减少while循环的次数，否则Leetcode会超时

广度优先 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";
            }
        }
    }
}

深度优先 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]));
                }
            }
        }
    }
}