[leetcode]Surrounded Regions @ Python

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<0 or x>m-1 or y<0 or y>n-1 or board[x][y]!='O':return
board[x][y] = 'D'
dfs(x-1, y)
dfs(x+1, y)
dfs(x, y+1)
dfs(x, y-1)

if len(board) == 0: return
m = len(board); n = len(board[0])
for i in range(m):
dfs(i, 0); dfs(i, n-1)
for j in range(1, n-1):
dfs(0, j); dfs(m-1, 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<0 or x>m-1 or y<0 or y>n-1 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(0); i=curr[0]; j=curr[1]
fill(i+1,j);fill(i-1,j);fill(i,j+1);fill(i,j-1)
if len(board)==0: return
m=len(board); n=len(board[0]); queue=[]
for i in range(n):
bfs(0,i); bfs(m-1,i)
for j in range(1, m-1):
bfs(j,0); bfs(j,n-1)
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'