LeetCode-Python-200. 島の数
'1'
(陸地)と'0'
(水)からなる2次元メッシュを与え,島の数を計算した.1つの島は水に囲まれ、水平方向または垂直方向に隣接する陸地で接続されている.グリッドの4つのエッジが水で囲まれていると仮定できます.例1:
:
11110
11010
11000
00000
: 1
例2:
:
11000
11000
00100
00011
: 3
第一の考え方:
DFS+visited配列は行ったポイントを保存します.
class Solution(object):
def numIslands(self, M):
"""
:type grid: List[List[str]]
:rtype: int
"""
if not M or not M[0]:
return 0
m, n = len(M), len(M[0])
visited = [[0 for j in range(n)] for i in range(m)]
# print visited
dx = [1, -1, 0, 0]
dy = [0, 0, 1, -1]
res = 0
def dfs(x0, y0):
for k in range(4):
x = x0 + dx[k]
y = y0 + dy[k]
# print x, y
if 0<= x < m and 0 <= y < n and M[x][y] == '1' and visited[x][y] ==0:
visited[x][y] = 1
dfs(x, y)
for i in range(m):
for j in range(n):
if M[i][j] == '1' and visited[i][j] == 0:
res += 1
visited[i][j] = 1
dfs(i, j)
# print visited
return res
2つ目の考え方:
DFS+染色、スキャン入力、「1」にスキャンすると、まずそれを「0」に変え、それにつながっている点を再帰的に「0」に変え、
この問題はLeetCodeのウェブサイトの出力が自動的にu"1"になる点で、解決方法は“1”と比べないで、“1”と比べるのです.encode(「utf-8」)をマッチングします.
# -*- coding: utf-8 -*-
class Solution(object):
def numIslands(self, grid):
"""
:type grid: List[List[str]]
:rtype: int
"""
island = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == "1".encode('utf-8'):
print i,j
island += 1
self.checkNearbyIsland(grid, i, j)
# print island
return island
def checkNearbyIsland(self, grid, i, j):
rows = len(grid)
columns = len(grid[0])
print i,j
if i > rows - 1 or j > columns - 1 or i < 0 or j < 0 or grid[i][j] == "0".encode('utf-8'):
return
if grid[i][j] == "1".encode('utf-8'):
grid[i][j] = "0"
# print grid
self.checkNearbyIsland(grid, i + 1, j)
self.checkNearbyIsland(grid, i, j + 1)
self.checkNearbyIsland(grid, i - 1, j)
self.checkNearbyIsland(grid, i, j - 1)
3つ目の考え方:
セットを使用して確認します.
# -*- coding: utf-8 -*-
class UnionFindSet(object):
def __init__(self, grid):
m, n = len(grid), len(grid[0])
self.roots = [-1 for i in range(m*n)]
self.rank = [0 for i in range(m*n)]
self.count = 0
for i in range(m):
for j in range(n):
if grid[i][j] == '1':
self.roots[i * n + j] = i * n + j
self.count += 1
def find(self, member):
tmp = []
while member != self.roots[member]:
tmp.append(member)
member = self.roots[member]
for root in tmp:
self.roots[root] = member
return member
def union(self, p, q):
parentP = self.find(p)
parentQ = self.find(q)
if parentP != parentQ:
if self.rank[parentP] > self.rank[parentQ]:
self.roots[parentQ] = parentP
elif self.rank[parentP] < self.rank[parentQ]:
self.roots[parentP] = parentQ
else:
self.roots[parentQ] = parentP
self.rank[parentP] -= 1
self.count -= 1
class Solution(object):
def numIslands(self, M):
"""
:type grid: List[List[str]]
:rtype: int
"""
if not M or not M[0]:
return 0
dx = [1, -1, 0, 0]
dy = [0, 0, 1, -1]
ufs = UnionFindSet(M)
m, n = len(M), len(M[0])
for i in range(m):
for j in range(n):
if M[i][j] == '1':
for k in range(4):
x = i + dx[k]
y = j + dy[k]
if 0 <= x < m and 0 <= y < n and M[x][y] == '1':
ufs.union(i*n + j, x *n + y)
# print ufs.roots
return ufs.count