477 被圍繞的區域

中文english

給乙個二維的矩陣，包含'x'和'o', 找到所有被'x'圍繞的區域，並用'x'替換其中所有的'o'。

樣例 1:

輸入:

x x x x
x o o x
x x o x
x o x x
輸出: 
x x x x
x x x x
x x x x
x o x x

輸入: 

x x x x
x o o x
x o o x
x o x x
輸出: 
x x x x
x o o x
x o o x
x o x x

輸入測試資料 (每行乙個引數)如何理解測試資料？

dfs + 邊界尋找o 替換 t

o替換x，t替換o

class

solution:
"""    @param: board: board a 2d board containing '
x' and 'o'
@return: nothing
"""def surroundedregions(self, board):
# write your code here
if not board: return
#初始化
n, m = len(board), len(board[0
])        def dfs(x, y):
if (x < 0 or x > n - 1 or y < 0 or y > m - 1
):                
return
if board[x][y] == 'o'
:                board[x][y] = 't'
dfs(x - 1
, y)
dfs(x + 1
, y)
dfs(x, y - 1
)                dfs(x, y + 1
)        #找尋邊界，如果滿足邊界 and 為0，則dfs找到上下左右的，換t
#此時j為0和m - 1
for i in
range(n):
if (board[i][0] == 'o'
):                dfs(i, 0)
if (board[i][m - 1] == 'o'
):                dfs(i, m - 1
)        
for j in
range(m):
if (board[0][j] == 'o'
):                dfs(
0, j)
if (board[n - 1][j] == 'o'
):                dfs(n - 1
, j)
#然後迴圈m和n，依次將o轉換為x，並且t轉換為o
for i in
range(n):
for j in
range(m):
if (board[i][j] == 'o'
):                    #board[i] = board[i].replace(board[i][j], '
x', 1
)                    board[i][j] = 'x'
elif (board[i][j] == 't'
):                    board[i][j] = 'o'
#board[i] = board[i].replace(board[i][j], '
o', 1
)        
return board

bfs寫法 + 邊界查詢 替換法

class

solution:
"""    @param: board: board a 2d board containing '
x' and 'o'
@return: nothing
"""def surroundedregions(self, board):
# write your code here
#bfs寫法，每次只fill一次，查詢一次,符合則替換，最終返回結果
if not board: return
m, n = len(board[0
]), len(board)
queue =
def fill(x, y):
if (x < 0 or x > n - 1 or y < 0 or y > m - 1
):                
return
if (board[x][y] == 'o'
):                board[x][y] = 't'
def bfs(x ,y):
fill(x, y)
while
queue:
curr = queue.pop(0
)                x, y = curr[0], curr[1
]                
fill(x + 1
, y)
fill(x - 1
, y)
fill(x, y + 1
)                fill(x, y - 1
)                
#邊界查詢
#x軸查詢
for i in
range(n):
if (board[i][0] == 'o'
):                bfs(i ,0)
if (board[i][m - 1] == 'o'
):                bfs(i, m - 1
)                
#y軸查詢
for j in
range(m):
if (board[0][j] == 'o'
):                bfs(
0, j)
if (board[n - 1][j] == 'o'
):                bfs(n - 1
, j)
#最終替換
for i in
range(m):
for j in
range(n):
if (board[j][i] == 'o'
):                    board[j][i] = 'x'
elif (board[j][i] == 't'
):                    board[j][i] = 'o'
return
board

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