Surrounded Regions----leetcode

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

题目的大意是指，将被'X'包围区域内的'o'替换为'X'，而对于未被包围的区域不用替换，可采用Dfs进行搜索边界上未被包围的'o',或者采用BFS的方法。

采用的java 递归的DFS方法出现了Runtime error，stack overflow，改为用栈实现递归。代码如下：

public class Solution {    public  void solve(char[][] board) {        if(board==null||board.length==0)            return;        int m,n;        m=board.length;        n=board[0].length;        for(int i=0;i<m;i++)        {            if(board[i][0]=='O') DFS(board,i,0);            if(board[i][n-1]=='O') DFS(board,i,n-1);        }        for(int j=1;j<n-1;j++)        {            if(board[0][j]=='O') DFS(board,0,j);            if(board[m-1][j]=='O') DFS(board,m-1,j);        }        for(int i=0;i<m;i++)            for(int j=0;j<n;j++)            {                if(board[i][j]=='O')                    board[i][j]='X';                else if(board[i][j]=='D')                    board[i][j]='O';            }    }    public void DFS(char[][] board,int x,int y)    {    if(x<0||x>=board.length||y<0||y>=board[0].length) return;    Stack<Integer> stack=new Stack<Integer>();        stack.push(x*board[0].length+y);        while(!stack.empty())        {            int curr=stack.pop();            int i=curr/board[0].length;            int j=curr%board[0].length;            board[i][j]='D';            if(i>0&&board[i-1][j]=='O')                 stack.push((i-1)*board[0].length+j);            if(i<board.length-1&&board[i+1][j]=='O')                stack.push((i+1)*board[0].length+j);            if(j>0&&board[i][j-1]=='O')                stack.push(i*board[0].length+j-1);            if(j<board[0].length-1&&board[i][j+1]=='O')                stack.push(i*board[0].length+j+1);        }    }}

采用BFS实现的代码如下，使用队列：

public class Solution {    private int m, n;    private char[][] board;    private Queue<Integer> queue = new LinkedList<Integer>();    public void add(int x, int y) {        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O') {            board[x][y] = 'Z';            queue.offer(x * n + y);        }    }        public void traversal(int x, int y) {        add(x, y);        while (!queue.isEmpty()) {            int p = queue.poll();            int px = p / n;            int py = p % n;            add(px - 1, py);            add(px + 1, py);            add(px, py - 1);            add(px, py + 1);        }    }    public void solve(char[][] board) {        // Start typing your Java solution below        // DO NOT write main() function        if (board == null || board.length == 0 || board[0].length == 0) {            return;        }        this.board = board;        m = board.length;        n = board[0].length;                for (int j = 0; j < n; j++) {            traversal(0, j);            traversal(m - 1, j);                    }        for (int i = 0; i < m; i++) {            traversal(i, 0);            traversal(i, n - 1);        }        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                board[i][j] = board[i][j] == 'Z' ? 'O' : 'X';            }        }    }}

两种方法在c++中实现问题不大，但是使用java会出现Runtime error或者TLE错误，检查是否发生越界、栈溢出或者死循环等问题。