【python/leetcode/130/M】Surrounded Regions
阿新 • • 發佈:2018-11-03
題目
https://leetcode.com/problems/surrounded-regions/
基本思路
轉換一下思路,找出哪些O是沒有被X包圍的。在面板四周的O肯定是沒有被X包圍的,與它們相連的O也是沒有被包圍的,其它的O都是被X包圍的。
問題簡化為將與四周的O相連的O都找出來,這些點不用變,其它點都變為X。
首先將四周的O壓入棧內,依次訪問棧內元素,並將它們標記,接著去判斷它們四周的元素是否也是O,如果是且沒有被標記過,則將其壓入棧中。當遍歷完棧中的元素後,將有標記的元素變為O,其餘都是X。
實現程式碼
class Solution(object):
def solve(self, board):
"""
:type board: List[List[str]]
:rtype: void Do not return anything, modify board in-place instead.
"""
if not board or not board[0]:
return
n,m = len(board),len(board[0])
# 找出所有的邊界O,將其押入棧中
stack = []
for i in range(n):
for j in range(m):
if ((i in (0, n - 1)) or (j in (0, m - 1))) and board[i][j] == 'O':
stack.append((i, j))
# 標記所有能聯絡到邊界的O的O
while stack:
r,c = stack.pop(0)
if 0<= r<n and 0<=c<m and board[r][c] == 'O':
board[r][c] = 'M'
if r-1>=0 and board[r-1][c] == 'O':
stack.append((r-1,c))
if r+1<n and board[r+1][c] == 'O':
stack.append((r+1,c))
if c-1>=0 and board[r][c-1] == 'O':
stack.append((r,c-1))
if c+1<m and board[r][c+1] == 'O':
stack.append((r,c+1))
# 更新
for i in range(n):
for j in range(m):
if board[i][j] == 'M':
board[i][j] = 'O'
else:
board[i][j] = 'X'