我为 DFS 非递归编写了一个解决方案，但我无法修改它来进行拓扑排序:

def dfs(graph,start):
    path = []
    stack = [start]    
    while stack != []: 
        v = stack.pop()
        if v not in path: path.append(v)
        for w in reversed(graph[v]): 
            if w not in path and not w in stack:
                stack.append(w) 
    return path

有什么修改方法吗？

使用递归版本，我可以轻松进行排序:

def dfs_rec(graph,start,path):
    path = path + [start]
    for edge in graph[start]: 
        if edge not in path:
            path = dfs_rec(graph, edge,path)
    print start
    return path

输入:

>>> graph = {
        1: [2, 3],
        2: [4, 5, 6],
        3: [4,6],
        4: [5,6],
        5: [6],
        6: []
    }
>>> dfs_rec(graph,1,[])
6
5
4
2
3
1
[1, 2, 4, 5, 6, 3]
>>> dfs(graph,1)
[1, 2, 4, 5, 6, 3]
>>> graph = {
        1: [3],
        3: [5,6],
        5: [4],
        4: [7],
        7: [],
        6: []
    }
>>> print dfs_rec(graph,1,[])
7
4
5
6
3
1
[1, 3, 5, 4, 7, 6]
>>> print dfs(graph,1)
[1, 3, 5, 4, 7, 6]

所以我也需要在非递归中得到这个排序。

非递归解决方案:

我认为这也可能是解决方案，如果我错了，请标记我。

def dfs(graph,start):
    path = []
    stack = [start]
    label = len(graph)
    result = {}  
    while stack != []:
        #this for loop could be done in other ways also
        for element in stack:
            if element not in result:
                result[element] = label
                label = label - 1

        v = stack.pop()
        if v not in path: path.append(v)
        for w in reversed(graph[v]): 
            if w not in path and not w in stack:
                stack.append(w) 

    result = {v:k for k, v in result.items()}
    return path,result

输入:

graph = { 1: [3], 3:[5,6] , 5:[4] , 4:[7], 7:[],6:[]}
print dfs(graph,1) 

输出:

([1, 3, 5, 4, 7, 6], {1: 7, 2: 4, 3: 5, 4: 6, 5: 3, 6: 1})

        1
       / 
      3
     /\
    5  6
   /
  4
 /
7    

最佳答案

FWIW，这是我为非递归拓扑排序编写的一些代码。

from collections import defaultdict, namedtuple
from itertools import islice

Results = namedtuple('Results', ['sorted', 'cyclic'])

def topological_sort(dependency_pairs):
    'Sort values subject to dependency constraints'
    num_heads = defaultdict(int)   # num arrows pointing in
    tails = defaultdict(list)      # list of arrows going out
    heads = []                     # unique list of heads in order first seen
    for h, t in dependency_pairs:
        num_heads[t] += 1
        if h in tails:
            tails[h].append(t)
        else:
            tails[h] = [t]
            heads.append(h)

    ordered = [h for h in heads if h not in num_heads]
    for h in ordered:
        for t in tails[h]:
            num_heads[t] -= 1
            if not num_heads[t]:
                ordered.append(t)
    cyclic = [n for n, heads in num_heads.items() if heads]
    return Results(ordered, cyclic)

if __name__ == '__main__':
    print( topological_sort('aa'.split()) )
    print( topological_sort('ah bg cf ch di ed fb fg hd he ib'.split()) )

关于python - 拓扑排序python，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/15038876/