题目：原题链接（中等）
标签：广度优先搜索、深度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N 2 ) | O ( N 2 ) | 324ms (16.72%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def shortestBridge(self, A: List[List[int]]) -> int:
m = n = len(A)
def _is_valid(x, y):
return 0 <= x < m and 0 <= y < n
def _get_neighbors(x1, y1):
return [(x2, y2) for (x2, y2) in [(x1 - 1, y1), (x1 + 1, y1), (x1, y1 - 1), (x1, y1 + 1)]
if _is_valid(x2, y2)]
island = []
visited = set()
for i in range(m):
for j in range(n):
if A[i][j] == 1 and (i, j) not in visited:
this = {(i, j)}
queue = collections.deque([(i, j)])
while queue:
i1, j1 = queue.popleft()
for (i2, j2) in _get_neighbors(i1, j1):
if A[i2][j2] == 1 and (i2, j2) not in this:
queue.append((i2, j2))
this.add((i2, j2))
island.append(set(this))
visited |= this
island1 = list(island[0])
island2 = island[1]
visited = set()
queue = collections.deque(island1)
step = 0
while queue:
for _ in range(len(queue)):
i1, j1 = queue.popleft()
for (i2, j2) in _get_neighbors(i1, j1):
if (i2, j2) in island2:
return step
if A[i2][j2] == 0 and (i2, j2) not in visited:
queue.append((i2, j2))
visited.add((i2, j2))
step += 1
