原题地址：https://oj.leetcode.com/problems/surrounded-regions/
题意：
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
解题思路：这道题可以使用BFS和DFS两种方法来解决。DFS会超时。BFS可以AC。从边上开始搜索，如果是'O'，那么搜索'O'周围的元素，并将'O'置换为'D'，这样每条边都DFS或者BFS一遍。而内部的'O'是不会改变的。这样下来，没有被围住的'O'全都被置换成了'D'，被围住的'O'还是'O'，没有改变。然后遍历一遍，将'O'置换为'X'，将'D'置换为'O'。
dfs代码，因为递归深度的问题会爆栈：
class Solution:
# @param board, a 9x9 2D array
# Capture all regions by modifying the input board in-place.
# Do not return any value.
def solve(self, board):
def dfs(x, y):
if x< or x>m- or y< or y>n- or board[x][y]!='O':return
board[x][y] = 'D'
dfs(x-, y)
dfs(x+, y)
dfs(x, y+)
dfs(x, y-) if len(board) == : return
m = len(board); n = len(board[])
for i in range(m):
dfs(i, ); dfs(i, n-)
for j in range(, n-):
dfs(, j); dfs(m-, j)
for i in range(m):
for j in range(n):
if board[i][j] == 'O': board[i][j] == 'X'
elif board[i][j] == 'D': board[i][j] == 'O'
bfs代码：
class Solution:
# @param board, a 9x9 2D array
# Capture all regions by modifying the input board in-place.
# Do not return any value.
def solve(self, board):
def fill(x, y):
if x< or x>m- or y< or y>n- or board[x][y] != 'O': return
queue.append((x,y))
board[x][y]='D'
def bfs(x, y):
if board[x][y]=='O':queue.append((x,y)); fill(x,y)
while queue:
curr=queue.pop(); i=curr[]; j=curr[]
fill(i+,j);fill(i-,j);fill(i,j+);fill(i,j-)
if len(board)==: return
m=len(board); n=len(board[]); queue=[]
for i in range(n):
bfs(,i); bfs(m-,i)
for j in range(, m-):
bfs(j,); bfs(j,n-)
for i in range(m):
for j in range(n):
if board[i][j] == 'D': board[i][j] = 'O'
elif board[i][j] == 'O': board[i][j] = 'X'