LeetCode. 130. 被围绕的区域

1、题目

https://leetcode-cn.com/problems/surrounded-regions/

2、题解

这是我学习到的一个思路，大概逻辑是
只要把不和边界O连通的 O 换位X就可以了。
所以我们只需要遍历一下四条边，将于边界O相连的O做上标记，最后遍历char[][]数组，来改变一下字母和符号就可以了。在找到边界O时，需要向它的四周去找和它相通的O，这里就用到了队列，被连通的O的坐标首先入队，然后出队，判断它的四周是否有连通的O，入队，再出队，判断是否有连通的O……直到队列中没有元素。

3、代码

class Solution {
        public void solve(char[][] board) {
            //排除异常
            if(board.length==0||board[0].length==0){
                return;
            }
            int row = board.length;
            int col = board[0].length;

            for (int i = 0; i < row; i++) {
                if (board[i][0] == 'O') helper(board,i,0);
                if (board[i][col-1] == 'O') helper(board,i,col-1);
            }

            for (int i = 0; i < col; i++) {
                if (board[0][i] == 'O') helper(board,0,i);
                if (board[row-1][i] == 'O') helper(board,row-1,i);
            }

            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] == '*') board[i][j] = 'O';
                }

        }

        public  void helper( char[][] board , int x, int y ){

            int[][] nums = {{0,1},{0,-1},{1,0},{-1,0}};
            Queue<Integer[]> queue = new LinkedList<>();
            Integer[] integers = {x,y};
            queue.offer(integers);
            board[x][y] = '*';

            while ( queue != null && queue.size() != 0){
                Integer[] temp = queue.poll();
                for (int i = 0; i < 4; i++) {
                    Integer newX = temp[0]+nums[i][0];
                    Integer newY = temp[1]+nums[i][1];
                    if (newX < 0 || newX >= board.length || newY < 0 || newY >= board[0].length) continue;

                    if (board[newX][newY] == 'O'){
                        board[newX][newY] = '*';
                        Integer[] t = {newX, newY};
                        queue.offer(t);
                    }
                }
            }

        }
    }

4、执行结果