深度优先搜索|130, 200
深度优先搜索|130. 被围绕的区域, 200. 岛屿数量
- 被围绕的区域
- 岛屿数量
被围绕的区域
这个题应该是从外到里做，我们应该去找和外圈的’O’相通的所有’O’，而不是去找内圈被包围的’O’。
所以我们做的就是从边界的’O’出发，把和他相连的所有’O’都标为’A’，也就是说遍历到最后，没有被接触过的（还是’O’）的那些就一定是被包围的。
那么把所有标为’A’的（没有被包围的’O’）转回’O’；然后把所有没被碰过的’O’（被包围的）转成’X’。
class Solution:
def solve(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
def direction(i,j,m,n):
l = [[i-1,j],[i+1,j],[i,j-1],[i,j+1]]
if i == 0:
l.remove([i-1,j])
if j == 0:
l.remove([i,j-1])
if i == m-1:
l.remove([i+1,j])
if j == n-1:
l.remove([i,j+1])
return l
m = len(board)
n = len(board[0])
used = [[False]*n for _ in range(m)]
def backtracking(i,j):
if board[i][j] != 'O': return
board[i][j] = 'A'
l = direction(i,j,m,n)
#print(i,j,l)
#print(board)
for k1,k2 in l:
backtracking(k1,k2)
for i in range(m):
backtracking(i,0)
backtracking(i,n-1)
for j in range(n):
backtracking(0,j)
backtracking(m-1,j)
for i in range(m):
for j in range(n):
if board[i][j] == 'A':
board[i][j] = 'O'
elif board[i][j] == 'O':
board[i][j] = 'X'
岛屿数量
和上一题一样，从边界的岛开始一路往里找连接的岛，而且题目说的很清晰。
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
def direction(i,j,m,n):
l = [[i-1,j],[i+1,j],[i,j-1],[i,j+1]]
if i == 0:
l.remove([i-1,j])
if j == 0:
l.remove([i,j-1])
if i == m-1:
l.remove([i+1,j])
if j == n-1:
l.remove([i,j+1])
return l
m = len(grid)
n = len(grid[0])
used = [[False]*n for _ in range(m)]
def backtracking(i,j):
if grid[i][j] == '0': return
used[i][j] = True
l = direction(i,j,m,n)
for k1,k2 in l:
if used[k1][k2]: continue
backtracking(k1,k2)
res = 0
for i in range(m):
for j in range(n):
if used[i][j] or grid[i][j] == '0': continue
backtracking(i,j)
res += 1
return res