Surrounded Regions

题目：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.
例子：

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X X X X
X O O X
X X O X
X O X X

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After running your function, the board should be:

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X X X X
X X X X
X X X X
X O X X

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方法1：要将所有以O组成、但不连通到网格边缘的区域变为X。所以我们可以先在四边上寻找连通到边缘的区域，将它们的O都变成Y。剩余的所有O一定无法连通到边缘，所以可以全部变为X。最后再将所有Y变回O。使用迭代版的DFS实现flood fill，当然也可以用BFS实现。

class Solution {public:    void solve(vector<vector<char>> &board) {        if(board.size()<3 || board[0].size()<3) return;        fillBorders(board, 'O', 'Y');        replace(board, 'O', 'X');        fillBorders(board, 'Y', 'O');    }    //迭代版的DFS    void fill(vector<vector<char>> &board, int i, int j, char target, char c) {        int m = board.size(), n = board[0].size();        if(i<0 || j<0 || i>=m || j>=n || board[i][j]!=target) return;        stack<pair<int,int>> s;        s.push(make_pair(i,j));        while(!s.empty()) {            i = s.top().first;            j = s.top().second;            s.pop();            board[i][j] = c;            if(i>0 && board[i-1][j]==target) s.push(make_pair(i-1,j));            if(i<m-1 && board[i+1][j]==target) s.push(make_pair(i+1,j));            if(j>0 && board[i][j-1]==target) s.push(make_pair(i,j-1));            if(j<n-1 && board[i][j+1]==target) s.push(make_pair(i,j+1));        }    }    void fillBorders(vector<vector<char>> &board, char target, char c) {        int m = board.size(), n = board[0].size();        for(int i=0; i<m; i++) {            if(board[i][0]==target) fill(board, i, 0, target, c);            if(board[i][n-1]==target) fill(board, i, n-1, target, c);        }        for(int j=1; j<n-1; j++) {            if(board[0][j]==target) fill(board, 0, j, target, c);            if(board[m-1][j]==target) fill(board, m-1, j, target, c);        }    }    void replace(vector<vector<char>> &board, char target, char c) {        int m = board.size(), n = board[0].size();        for(int i=0; i<m; i++) {            for(int j=0; j<n; j++) {                if(board[i][j]==target)                    board[i][j] = c;            }        }    }};

方法2：BFS

public class Solution {    static final int[] directionX = {+1, -1, 0, 0};    static final int[] directionY = {0, 0, +1, -1};    static final char FREE = 'F';    static final char TRAVELED = 'T';    public void solve(char[][] board) {        if (board.length == 0) {            return;        }        int row = board.length;        int col = board[0].length;        for (int i = 0; i < row; i++) {            bfs(board, i, 0);            bfs(board, i, col - 1);        }        for (int j = 1; j < col - 1; j++) {            bfs(board, 0, j);            bfs(board, row - 1, j);        }        for (int i = 0; i < row; i++) {            for (int j = 0; j < col; j++) {                switch(board[i][j]) {                    case 'O':                         board[i][j] = 'X';                        break;                    case 'F':                        board[i][j] = 'O';                }            }        }    }    public void bfs(char[][] board, int i, int j) {        if (board[i][j] != 'O') {            return;        }        Queue<Node> queue = new LinkedList<Node>();        queue.offer(new Node(i, j));        while (!queue.isEmpty()) {            Node crt = queue.poll();            board[crt.x][crt.y] = FREE;            for (Node node : expand(board, crt)) {                queue.offer(node);            }        }    }    private List<Node> expand(char[][] board, Node node) {        List<Node> expansion = new ArrayList<Node>();        for (int i = 0; i < directionX.length; i++) {            int x = node.x + directionX[i];            int y = node.y + directionY[i];            // check validity            if (x >= 0 && x < board.length && y >= 0 && y < board[0].length && board[x][y] == 'O') {                board[x][y] = TRAVELED;                expansion.add(new Node(x, y));            }        }        return expansion;    }    static class Node {        int x;        int y;        Node(int x, int y) {            this.x = x;            this.y = y;        }    }}

方法3：（九章算法上题解，没理解代码）

public class Solution {    private static Queue<Integer> queue = null;    private static char[][] board;    private static int rows = 0;    private static int cols = 0;    public void solve(char[][] board) {        // Note: The Solution object is instantiated only once and is reused by each test case.        if (board.length == 0 || board[0].length == 0) return;        queue = new LinkedList<Integer>();        board = board;        rows = board.length;        cols = board[0].length;        for (int i = 0; i < rows; i++) { // **important**            enqueue(i, 0);            enqueue(i, cols - 1);        }        for (int j = 1; j < cols - 1; j++) { // **important**            enqueue(0, j);            enqueue(rows - 1, j);        }        while (!queue.isEmpty()) {            int cur = queue.poll();            int x = cur / cols,                y = cur % cols;            if (board[x][y] == 'O') {                board[x][y] = 'D';            }            enqueue(x - 1, y);            enqueue(x + 1, y);            enqueue(x, y - 1);            enqueue(x, y + 1);        }        for (int i = 0; i < rows; i++) {            for (int j = 0; j < cols; j++) {                if (board[i][j] == 'D') board[i][j] = 'O';                else if (board[i][j] == 'O') board[i][j] = 'X';            }        }        queue = null;        board = null;        rows = 0;        cols = 0;    }    public static void enqueue(int x, int y) {        if (x >= 0 && x < rows && y >= 0 && y < cols && board[x][y] == 'O'){              queue.offer(x * cols + y);        }    }}