我编写了深度优先搜索,返回找到目标节点的深度,如果没有找到路径则返回-1。算法有效,但我需要加快速度。这是函数
def depth(dic, head, target):
if(head==target):
return
depth=1
que = deque()
que.append('|') #used to mark end of each breadth so i can count depth correctly
used = list()
add = True
while(que): #while the que isn't empty
for x in dic[head]: #check current level
if x==target:
print(depth),
return;
for x in dic[head]: #add this level to the que and used list
for y in used:
if y==x:
add=False
break
if add == True:
que.append(x)
used.append(x)
add=True
que.append('|') #add our delimiter
while(que): #grab the next item from the que and inc depth count
temp = que.popleft()
if temp=='|': #bump depth counter since we found end of a level
depth+=1
else:
head=temp #bump to next node to check
break
print('-1'), #reached the end, node not found
传入的dic被声明为
dic = defaultdict(list)
这样每个值都是一个整数列表,并且我使用了|所以我知道何时碰到深度计数器。我意识到我在中间陷入困境,在那里我检查当前级别的所有节点并将它们添加到que和used list中,但我不知道如何加快速度。
编辑:
对于任何有类似问题的人来说,这是我最终得到的算法,它通过搜索的方式有点奇怪,因为我返回了可以找到值的最浅深度,如果并非同时检查同一深度的所有连接,我们最终可能会在下一个深度找到节点(如一个错误的关闭)
def depthOfNodeBFS(dic, head, target):
if(head==target):
return
depth=0
que = []
que.append(head)
used = set()
nex = list()
while(que): #while the que isn't empty
depth+=1 #bump the depth counter
for x in que: #check the next level of all nodes at current depth
if target in dic[x]: #if the target is found were done
print(depth),
return;
else: #other wise track what we've checked
nex.append(x)
used.add(x)
while(nex): #remove checked from que and add children to que
que.pop(0)
que.extend(dic[nex.pop()]-used)
print('-1'),
答案 0 :(得分:2)
这看起来更像是广度优先搜索而不是深度优先搜索给我,但你不应该有嵌套的while
循环。广度优先搜索的常用算法类似于:
add the root to the queue
while the queue is not empty:
dequeue the first element elt
if elt is the solution, return it
otherwise, add each child of elt to the queue
如果你想报告深度,我建议在队列中添加元组(节点,深度):
add (root, 0) to the queue
while the queue is not empty:
elt, depth = queue.popleft()
if elt is the solution, return (elt, depth)
otherwise, for each child of elt:
add (child, depth+1) to the queue