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
从四个边入手，把不能换成X的O标记出来，用map存。 然后再遍历这个array， 更改可以变成X的。
1 public class Solution { 2 public void solve(char[][] board) { 3 // Start typing your Java solution below 4 // DO NOT write main() function 5 if(board == null || board.length == 0 || board[0].length == 0) return; 6 boolean[][] map = new boolean[board.length][board[0].length]; 7 int len = board[0].length; 8 LinkedList<Integer> ll = new LinkedList<Integer>(); 9 for(int i = 0; i < board.length; i ++){ 10 if(board[i][0] == 'O' && !map[i][0]){ 11 map[i][0] = true; 12 ll.push(i * len); 13 } 14 if(board[i][len - 1] == 'O' && !map[i][len - 1]){ 15 map[i][len - 1] = true; 16 ll.push(i * len + len - 1); 17 } 18 } 19 for(int i = 0; i < len; i ++){ 20 if(board[0][i] == 'O' && !map[0][i]){ 21 map[0][i] = true; 22 ll.push(i); 23 } 24 if(board[board.length - 1][i] == 'O' && !map[board.length - 1][i]){ 25 map[board.length - 1][i] = true; 26 ll.push((board.length - 1) * len + i); 27 } 28 } 29 while(ll.size() != 0){ 30 int tmp = ll.poll(); 31 int i = tmp / len; 32 int j = tmp % len; 33 if(i > 1 && board[i - 1][j] == 'O' && !map[i - 1][j]){ 34 map[i - 1][j] = true; 35 ll.push((i - 1) * len + j); 36 } 37 if(j > 1 && board[i][j - 1] == 'O' && !map[i][j - 1]){ 38 map[i][j - 1] = true; 39 ll.push(i * len + j - 1); 40 } 41 if(i + 1 < board.length && board[i + 1][j] == 'O' && !map[i + 1][j]){ 42 map[i + 1][j] = true; 43 ll.push((i + 1) * len + j); 44 } 45 if(j + 1 < len && board[i][j + 1] == 'O' && !map[i][j + 1]){ 46 map[i][j + 1] = true; 47 ll.push(i * len + j + 1); 48 } 49 } 50 for(int i = 0; i < board.length; i ++) 51 for(int j = 0; j < len; j ++) 52 if(!map[i][j] && board[i][j] == 'O') 53 board[i][j] = 'X'; 54 } 55 }