Surrounded Regions (Java)

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

这道题的包围是指上下左右均被x包围的情况，多个O时，是一起被包围，而不是一个O被X消灭后再消灭

比如

X X X X

X O O X

X X X X

两个O外围全是X所以被消灭。

如果遍历时改变O为X，可能会影响其他O是否被消灭。

这道题可以从四周边界处开始考虑，凡是与边界O相连的O都不会被X消灭。

所以从边界O开始进行DFS或BFS。

起初用的DFS写的，java的DFS在这道题中会出现栈溢出。之后用BFS写的可以过。注意每次是将坐标入队列。

Source

public class Solution {    public void solve(char[][] board) {    if(board.length == 0) return;           for(int i = 0; i < board.length; i++){        bfs(i, 0, board);    bfs(i, board[0].length - 1, board);    }       for(int j = 0; j < board[0].length; j++){    bfs(0, j, board);    bfs(board.length - 1, j, board);    }       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';    }    else if(board[i][j] == '@'){    board[i][j] = 'O';    }    }    }       return;    }    public void bfs(int i, int j, char[][] board){    if(board[i][j] == 'X') return;    if(board[i][j] == '@') return;   LinkedList<Integer> queue = new LinkedList<Integer>();      int len = board[0].length;   board[i][j] = '@';   int a = i * len + j; //queue里面要存的是坐标 这样才能在BFS的时候传递坐标关系   queue.add(a);   while(!queue.isEmpty()){   a = queue.poll();   int r = a / len;   int c = a % len;      if(r - 1 >= 0 && board[r - 1][c] == 'O'){   board[r - 1][c] = '@';   queue.add((r - 1) * len + c);    }   if(r + 1 < board.length && board[r + 1][c] == 'O'){   board[r + 1][c] = '@';   queue.add((r + 1) * len + c);   }   if(c - 1 >= 0 && board[r][c - 1] == 'O'){   board[r][c - 1] = '@';   queue.add(r * len + c - 1);   }   if(c + 1 < len && board[r][c + 1] == 'O'){   board[r][c + 1] = '@';   queue.add(r * len + c + 1);   }   }          }}

Test

    public static void main(String[] args){    char[][] board = {{'X','O','X','X'},{'O','X','O','X'},{'X','O','X','O'},{'O','X','O','X'},{'X','O','X','O'},{'O','X','O','X'}};    new Solution().solve(board);    for(int i = 0; i < board.length; i++){    for(int j = 0; j < board[0].length; j++){    System.out.print(board[i][j]);    System.out.print(" ");    }    System.out.println();    }    }