首先，你并不是真的想要找出所有的循环因为如果有1个，那么就会有无穷多个循环。比如A-B-A, A-B-A- b - a等等。或者可以将2个循环组合成一个8-like循环等等……有意义的方法是寻找所有所谓的简单循环——那些除了开始/结束点之外不交叉的循环。如果你愿意，你可以生成简单循环的组合。

One of the baseline algorithms for finding all simple cycles in a directed graph is this: Do a depth-first traversal of all simple paths (those that do not cross themselves) in the graph. Every time when the current node has a successor on the stack a simple cycle is discovered. It consists of the elements on the stack starting with the identified successor and ending with the top of the stack. Depth first traversal of all simple paths is similar to depth first search but you do not mark/record visited nodes other than those currently on the stack as stop points.

The brute force algorithm above is terribly inefficient and in addition to that generates multiple copies of the cycles. It is however the starting point of multiple practical algorithms which apply various enhancements in order to improve performance and avoid cycle duplication. I was surprised to find out some time ago that these algorithms are not readily available in textbooks and on the web. So I did some research and implemented 4 such algorithms and 1 algorithm for cycles in undirected graphs in an open source Java library here : http://code.google.com/p/niographs/ .

顺便说一句，因为我提到了无向图:它们的算法是不同的。构建一棵生成树，然后每一条不属于树的边与树中的一些边一起形成一个简单的循环。这样发现的循环形成了所谓的循环基。所有的简单循环都可以通过组合两个或多个不同的基循环来找到。更多细节请参见:http://dspace.mit.edu/bitstream/handle/1721.1/68106/FTL_R_1982_07.pdf。

在无向图的情况下，最近发表的一篇论文(无向图中循环和st路径的最优列表)提供了一个渐近最优解。你可以在这里阅读http://arxiv.org/abs/1205.2766或http://dl.acm.org/citation.cfm?id=2627951 我知道它不能回答你的问题，但由于你的问题标题没有提到方向，它可能仍然对谷歌搜索有用

深度优先搜索和回溯应该在这里工作。 保存一个布尔值数组，以跟踪您以前是否访问过某个节点。如果您没有新节点可访问(不涉及已经访问过的节点)，那么只需返回并尝试不同的分支。

如果你有一个邻接表来表示图，DFS很容易实现。例如adj[A] = {B,C}表示B和C是A的子结点。

例如，下面的伪代码。“start”是开始的节点。

dfs(adj,node,visited):  
  if (visited[node]):  
    if (node == start):  
      "found a path"  
    return;  
  visited[node]=YES;  
  for child in adj[node]:  
    dfs(adj,child,visited)
  visited[node]=NO;

用开始节点调用上面的函数:

visited = {}
dfs(adj,start,visited)

基于dfs的带有后边缘的变体确实会发现循环，但在许多情况下，它不会是最小循环。一般来说，DFS给出了存在循环的标志，但它不足以真正找到循环。例如，想象5个不同的循环共用两条边。仅仅使用DFS(包括回溯变量)没有简单的方法来识别周期。

Johnson算法确实给出了所有唯一的简单循环，并具有良好的时间和空间复杂度。

但如果你只想找到最小循环(意味着可能有多个循环通过任何顶点，我们感兴趣的是找到最小循环)，并且你的图不是很大，你可以尝试使用下面的简单方法。 它非常简单，但与约翰逊的相比相当慢。

So, one of the absolutely easiest way to find MINIMAL cycles is to use Floyd's algorithm to find minimal paths between all the vertices using adjacency matrix. This algorithm is nowhere near as optimal as Johnson's, but it is so simple and its inner loop is so tight that for smaller graphs (<=50-100 nodes) it absolutely makes sense to use it. Time complexity is O(n^3), space complexity O(n^2) if you use parent tracking and O(1) if you don't. First of all let's find the answer to the question if there is a cycle. The algorithm is dead-simple. Below is snippet in Scala.

  val NO_EDGE = Integer.MAX_VALUE / 2

  def shortestPath(weights: Array[Array[Int]]) = {
    for (k <- weights.indices;
         i <- weights.indices;
         j <- weights.indices) {
      val throughK = weights(i)(k) + weights(k)(j)
      if (throughK < weights(i)(j)) {
        weights(i)(j) = throughK
      }
    }
  }

Originally this algorithm operates on weighted-edge graph to find all shortest paths between all pairs of nodes (hence the weights argument). For it to work correctly you need to provide 1 if there is a directed edge between the nodes or NO_EDGE otherwise. After algorithm executes, you can check the main diagonal, if there are values less then NO_EDGE than this node participates in a cycle of length equal to the value. Every other node of the same cycle will have the same value (on the main diagonal).

为了重建周期本身，我们需要使用带有父跟踪的稍微修改版本的算法。

  def shortestPath(weights: Array[Array[Int]], parents: Array[Array[Int]]) = {
    for (k <- weights.indices;
         i <- weights.indices;
         j <- weights.indices) {
      val throughK = weights(i)(k) + weights(k)(j)
      if (throughK < weights(i)(j)) {
        parents(i)(j) = k
        weights(i)(j) = throughK
      }
    }
  }

如果顶点之间有边，父矩阵最初应该包含边缘单元中的源顶点索引，否则为-1。 函数返回后，对于每条边，您都将引用到最短路径树中的父节点。 然后很容易恢复实际的循环。

总之，我们有下面的程序来求所有的最小循环

  val NO_EDGE = Integer.MAX_VALUE / 2;

  def shortestPathWithParentTracking(
         weights: Array[Array[Int]],
         parents: Array[Array[Int]]) = {
    for (k <- weights.indices;
         i <- weights.indices;
         j <- weights.indices) {
      val throughK = weights(i)(k) + weights(k)(j)
      if (throughK < weights(i)(j)) {
        parents(i)(j) = parents(i)(k)
        weights(i)(j) = throughK
      }
    }
  }

  def recoverCycles(
         cycleNodes: Seq[Int], 
         parents: Array[Array[Int]]): Set[Seq[Int]] = {
    val res = new mutable.HashSet[Seq[Int]]()
    for (node <- cycleNodes) {
      var cycle = new mutable.ArrayBuffer[Int]()
      cycle += node
      var other = parents(node)(node)
      do {
        cycle += other
        other = parents(other)(node)
      } while(other != node)
      res += cycle.sorted
    }
    res.toSet
  }

还有一个小的main方法来测试结果

  def main(args: Array[String]): Unit = {
    val n = 3
    val weights = Array(Array(NO_EDGE, 1, NO_EDGE), Array(NO_EDGE, NO_EDGE, 1), Array(1, NO_EDGE, NO_EDGE))
    val parents = Array(Array(-1, 1, -1), Array(-1, -1, 2), Array(0, -1, -1))
    shortestPathWithParentTracking(weights, parents)
    val cycleNodes = parents.indices.filter(i => parents(i)(i) < NO_EDGE)
    val cycles: Set[Seq[Int]] = recoverCycles(cycleNodes, parents)
    println("The following minimal cycle found:")
    cycles.foreach(c => println(c.mkString))
    println(s"Total: ${cycles.size} cycle found")
  }

输出是

The following minimal cycle found:
012
Total: 1 cycle found

我发现解决这个问题的最简单的选择是使用名为networkx的python库。

它实现了这个问题的最佳答案中提到的约翰逊算法，但它的执行非常简单。

简而言之，你需要以下几点:

import networkx as nx
import matplotlib.pyplot as plt

# Create Directed Graph
G=nx.DiGraph()

# Add a list of nodes:
G.add_nodes_from(["a","b","c","d","e"])

# Add a list of edges:
G.add_edges_from([("a","b"),("b","c"), ("c","a"), ("b","d"), ("d","e"), ("e","a")])

#Return a list of cycles described as a list o nodes
list(nx.simple_cycles(G))

答案:[['a'， 'b'， 'd'， 'e']， ['a'， 'b'， 'c']]

从节点X开始，检查所有子节点(如果无方向，父节点和子节点是等价的)。将这些子节点标记为X的子节点。对于任何这样的子节点A，标记它的子节点是A的子节点，X'，其中X'标记为2步远。)如果您稍后点击X并将其标记为X的子节点”，这意味着X处于3节点周期中。回溯到它的父节点很容易(因为算法不支持这一点，所以你可以找到任何一个有X'的父节点)。

注意:如果图是无向的或者有任何双向边，这个算法会变得更复杂，假设你不想在一个周期内两次遍历同一条边。