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
反向思索最简单：哪些‘O’是应该保留的？
? 从上下左右四个边界往里走，凡是能碰到的
‘O’ ，都是跟边界接壤的，应该保留。
? 思路：
? 对于每一个边界上的‘O’作为起点，做若干次广度
优先搜索，对于碰到的‘O’，标记为其他某字符Y；
? 最后遍历一遍整个地图，把所有的Y恢复成‘O’，把
所有现有的‘O’都改成‘X’
1 import Queue 2 class Solution(object): 3 def solve(self, board): 4 """ 5 :type board: List[List[str]] 6 :rtype: void Do not return anything, modify board in-place instead. 7 """ 8 def fill(x, y): 9 if x < 0 or x > height-1 or y < 0 or y > width-1 or board[x][y] != "O": 10 return 11 MyQueue.put((x, y)) 12 board[x][y] = "D" 13 14 def bfs(x, y): 15 if board[x][y] == "O": 16 fill(x, y) 17 18 while not MyQueue.empty(): 19 current = MyQueue.get() 20 i, j = current[0], current[1] 21 fill(i+1, j) 22 fill(i-1, j) 23 fill(i, j+1) 24 fill(i, j-1) 25 26 if len(board) == 0: 27 return 28 29 height, width, MyQueue = len(board), len(board[0]), Queue.Queue() 30 for i in range(width): 31 bfs(0, i) 32 bfs(height - 1, i) 33 34 for j in range(1, height - 1): 35 bfs(j, 0) 36 bfs(j, width - 1) 37 38 for i in range(height): 39 for j in range(width): 40 if board[i][j] == "D": 41 board[i][j] = "O" 42 elif board[i][j] == "O": 43 board[i][j] = "X"
参考：https://www.jianshu.com/p/3ea288ffdb68
