leetcode 130. Surrounded Regions----- java

Given a 2D board containing 'X' and 'O' (the letter 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

像围棋一样，去掉被包围的O，刚开始用了HashSet<Integer>但是内存溢出了。

public class Solution {
    public void solve(char[][] board) {
        int row = board.length;
        if( row < 2 )
            return ;
        int col = board[0].length;
        if( col < 2 )
            return ;
        
        for( int i = 0;i<row;i++){
            for( int j = 0;j<col;j++){
                if( board[i][j] == 'O'){
                    HashSet<Integer> set = new HashSet<Integer>();
                    if( helper(board,i,j,set,false) ){
                        helpset('o',set,board);
                    }else
                        helpset('X',set,board);
                }
            }
        }
        for( int i = 0;i<row;i++){
            for( int j = 0;j< col;j++){
                if( board[i][j] == 'o')
                    board[i][j] = 'O';
            }
        }
    }

    public void helpset(char ch ,HashSet<Integer> set ,char[][] board){

        for( int i : set ){
            int row = i/board[0].length;
            int col = i%board[0].length;
            board[row][col] = ch;
        }
    }


    public boolean helper(char[][] board,int num1, int num2,HashSet<Integer> set,boolean flag){
        boolean result = flag;
        if( num1-1 >= 0 && board[num1-1][num2] == 'O'){
            int num = (num1-1)*board[0].length+num2;
            if( !set.contains(num) ){
                set.add(num);
                result = result || helper(board,num1-1,num2,set,flag);
            }
        }
        if( num1+1 < board.length && board[num1+1][num2] == 'O'){
            int num = (num1+1)*board[0].length+num2;
            if( !set.contains(num) ){
                set.add(num);
                result = result || helper(board,num1+1,num2,set,flag);
            }
        }
        if( num2-1 >= 0 && board[num1][num2-1] == 'O'){
            int num = num1*board[0].length+num2-1;
            if( !set.contains(num) ){
                set.add(num);
                result = result || helper(board,num1,num2-1,set,flag);
            }
        }
        if( num2+1 < board[0].length && board[num1][num2+1] == 'O'){
            int num = num1*board[0].length+num2+1;
            if( !set.contains(num) ){
                set.add(num);
                result = result || helper(board,num1,num2+1,set,flag);
            }
        }

        if( num1 == 0 || num1 == board.length-1 || num2 == 0 || num2 == board[0].length-1)
            return true;


        return result;

    }
}

 2、换用思路，先把所有的O放入一个set中，然后在这个set中进行判断，虽然ac但是还是耗时还是较长。

public class Solution {
    public void solve(char[][] board) {

        int row = board.length;
        if( row < 2 )
            return ;
        int col = board[0].length;
        if( col < 2 )
            return ;
        HashSet<Integer> set = new HashSet<Integer>();
        for( int i = 0;i<row;i++){
            for( int j = 0;j<col;j++){
                if( board[i][j] == 'O')
                    set.add(i*board[0].length+j);
            }
        }
        while( !set.isEmpty() ){
            helper(set,col,row,board);
        }
        return ;
    }
    public void helper(HashSet<Integer> set,int len,int row,char[][] board){

         boolean result = false;
        Queue<Integer> queue = new LinkedList<Integer>();
        int[] set2 = new int[set.size()];
        int num = set.iterator().next();
        set2[0] = num;
        int i = 1;
        queue.add(num);
        set.remove(num);
        if (num % len == 0 || num % len == len - 1 || num / len == 0 || num / len == row - 1)
            result = true;
        while( !queue.isEmpty() ){
            num = (Integer) queue.poll();
            if( set.contains(num+1) ){
                set2[i] = num+1;
                i++;
                queue.add(num+1);
                set.remove(num+1);
                if ((num+1) % len == 0 || (num+1) % len == len - 1 || (num+1) / len == 0 || (num+1) / len == row - 1)
                    result = true;
            }
            if( set.contains(num-1)){
                set2[i] = (num-1);
                i++;
                queue.add(num-1);
                set.remove(num-1);
                if ((num-1) % len == 0 ||( num-1) % len == len - 1 || (num-1) / len == 0 || (num-1) / len == row - 1)
                    result = true;
            }
            if( set.contains(num+len) ){
                set2[i] = (num+len);
                i++;
                queue.add(num+len);
                set.remove(num+len);
                if ((num+len) % len == 0 || (num+len )% len == len - 1 || (num+len) / len == 0 || (num+len) / len == row - 1)
                    result = true;
            }
            if( set.contains(num-len) ){
                set2[i] = (num-len);
                i++;
                queue.add(num-len);
                set.remove(num-len);
                if ((num-len) % len == 0 || (num-len) % len == len - 1 || (num-len) / len == 0 || (num-len) / len == row - 1)
                   result = true;
            }
        }
        if( result == true )
            return ;
        else
            helpset('X',set2,i,board);
    }
    public void helpset(char ch ,int[] set,int num ,char[][] board){
        for( int i = 0;i<num;i++){
            int row = set[i]/board[0].length;
            int col = set[i]%board[0].length;
            board[row][col] = ch;
        }
    }
}

 3、使用BFS和队列。

public class Solution {
    public void solve(char[][] board) {
        int row = board.length;
        if( row < 2 )
            return ;
        int col = board[0].length;
        if( col < 2 )
            return ;
        Queue<Integer> queue = new LinkedList<Integer>();
        for( int i = 0;i<col;i++){
            if( board[0][i] == 'O' )
                queue.add(i);
            if( board[row-1][i] == 'O')
                queue.add((row-1)*col+i);
        }
        for( int i = 0;i < row ;i++){
            if( board[i][col-1] == 'O')
                queue.add(i*col+col-1);
            if( board[i][0] == 'O')
                queue.add(i*col);
        }
        while( !queue.isEmpty() ){
            int num = queue.poll();
            int x = num/col,y = num%col;
            if( board[x][y] != 'O')
                continue;
            board[x][y] = 'o';
            if( x-1>=0 &&  board[x-1][y] == 'O')
                queue.add(num-col);
            if( x+1<row && board[x+1][y] == 'O')
                queue.add(num+col);
            if( y-1>=0 && board[x][y-1] == 'O')
                queue.add(num-1);
            if( y+1<col && board[x][y+1] == 'O')
                queue.add(num+1);
        }
        
        for( int i = 0;i<row;i++){
            for( int j = 0;j<col;j++){
                if( board[i][j] == 'O')
                    board[i][j] = 'X';
                else if( board[i][j] == 'o')
                    board[i][j] = 'O';
            }
        }



        return ;
    }
}

4、不使用队列，直接用BFS。

public class Solution {
    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++) {
            if (board[i][0] == 'O')
                dfs(board, i, 0);
            if (board[i][col-1] == 'O')
                dfs(board, i, col-1);
        }
        
        for (int i = 1; i < col - 1; i++) {
            if (board[0][i] == 'O')
                dfs(board, 0, i);
            if (board[row - 1][i] == 'O')
                dfs(board,row-1, i);
        }
        
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == '1') {
                    board[i][j] = 'O';
                } else {
                    board[i][j] = 'X';
                }
            }
        }
        return;
        
    }
    public void dfs(char[][] board, int m, int n) {
        if (board[m][n] != 'O') return;
        board[m][n] = '1';
        if (m < board.length - 2)
            dfs(board, m + 1, n);
        if (m > 1)
            dfs(board, m - 1, n);
        if (n < board[0].length - 2)
            dfs(board, m, n + 1);
        if (n > 1)
            dfs(board, m, n - 1);
        
    }
}