题目：原题链接（中等）
标签：图、广度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( N ) | 132ms (24.37%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
def build_graph(edges):
graph = collections.defaultdict(set)
for edge in edges:
graph[edge[0]].add(edge[1])
return graph
class Solution:
def shortestAlternatingPaths(self, n: int, red_edges: List[List[int]], blue_edges: List[List[int]]) -> List[int]:
red_graph = build_graph(red_edges)
blue_graph = build_graph(blue_edges)
ans = [-1] * n
# 先红色边的情况
visited_red = {0}
visited_blue = set()
queue = collections.deque([0])
step = 0
while queue:
for _ in range(len(queue)):
n1 = queue.popleft()
if ans[n1] == -1 or ans[n1] > step:
ans[n1] = step
if step % 2 == 0:
for n2 in red_graph[n1]:
if n2 not in visited_blue:
visited_blue.add(n2)
queue.append(n2)
else:
for n2 in blue_graph[n1]:
if n2 not in visited_red:
visited_red.add(n2)
queue.append(n2)
step += 1
# 先蓝色边的情况
visited_red = set()
visited_blue = {0}
queue = collections.deque([0])
step = 0
while queue:
for _ in range(len(queue)):
n1 = queue.popleft()
if ans[n1] == -1 or ans[n1] > step:
ans[n1] = step
if step % 2 == 0:
for n2 in blue_graph[n1]:
if n2 not in visited_red:
visited_red.add(n2)
queue.append(n2)
else:
for n2 in red_graph[n1]:
if n2 not in visited_blue:
visited_blue.add(n2)
queue.append(n2)
step += 1
return ans
