题目：原题链接（中等）
标签：图、拓扑排序、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 ) O(N) O(N) | O ( N ) O(N) O(N) | 84ms (12.88%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
def build_graph(edges):
graph_in = collections.defaultdict(set)
graph_out = collections.defaultdict(set)
for edge in edges:
graph_in[edge[1]].add(edge[0])
graph_out[edge[0]].add(edge[1])
return graph_out, graph_in
def topo(graph_in, graph_out):
count = {} # 节点入射边统计
queue = [] # 当前入射边为0的节点列表
# 统计所有节点的入射边
for node in graph_in:
count[node] = len(graph_in[node])
for node in graph_out:
if node not in count:
count[node] = 0
queue.append(node)
# 拓扑排序
order = []
while queue:
node = queue.pop()
order.append(node)
for next in graph_out[node]:
count[next] -= 1
if count[next] == 0:
queue.append(next)
return order
class Solution:
def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
# 生成有向图中边的邻接列表结构
graph_in, graph_out = build_graph(prerequisites)
# 拓扑排序
order = topo(graph_in, graph_out)
order_set = set(order)
for node in graph_in:
if node not in order:
return []
for i in range(numCourses):
if i not in order_set:
order.append(i)
return order
