[Leetcode] 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

从外围往里找，若存在都为‘O’的路径，则不变，不存在则变

public class Solution {    public void solve(char[][] board) {if (board.length == 0 || board[0].length == 0)return;for (int i = 0; i < board.length; i++) {if (board[i][0] == 'O')board[i][0] = 'A';if (board[i][board[0].length - 1] == 'O')board[i][board[0].length - 1] = 'A';}for (int j = 0; j < board[0].length; j++) {if (board[0][j] == 'O')board[0][j] = 'A';if (board[board.length - 1][j] == 'O')board[board.length - 1][j] = 'A';}for (int i = 1; i < board.length - 1; i++) {for (int j = 1; j < board[0].length - 1; j++) {if (board[i][j] == 'O'&& (board[i - 1][j] == 'A' || board[i][j - 1] == 'A'))board[i][j] = 'A';}}for (int i = board.length - 2; i > 0; i--) {for (int j = board[0].length - 2; j > 0; j--) {if (board[i][j] == 'O'&& (board[i + 1][j] == 'A' || board[i][j + 1] == 'A'))board[i][j] = 'A';}}int time = board.length < board[0].length ? board.length: board[0].length;for (int k = 1; k < time; ++k) {for (int i = 1; i < board.length; ++i) {for (int j = 1; j < board[0].length; ++j) {if (board[i][j] == 'O') {if (board[i][j - 1] == 'A' || board[i - 1][j] == 'A')board[i][j] = 'A';if (j + 1 < board[0].length && board[i][j + 1] == 'A')board[i][j] = 'A';if (i + 1 < board.length && board[i + 1][j] == 'A')board[i][j] = 'A';}}}}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';if (board[i][j] == 'A')board[i][j] = 'O';}}}}