Google – Print All Cycles in Graph


Google – Print All Cycles in Graph
很简单的题,但是对于是否会出现无限循环?或者发现cycle后要不要把visit[]置1(表示已经结束)?这两个想来想去想不清楚的问题,赤裸裸的证明了我recursion和backtracking根本就没想明白过….
首先会不会出现无限循环?
答案是no. 不会出现无限循环. 因为如果找到了一个cycle,也就是回到了某个visit过但是还没有结束的vertex,我们做的是输出cycle后return,而不是继续往下找,如果继续往下找就会出现无限循环了。所以无论图里有几个cycle,都不会出现无限循环的情况。
其次要不要置1?
答案也是no. 找到cycle后不能把visit[]置1!除非题目假设了graph里只有一个cycle,否则想象一下有一个点,从这个点出发,左边有一个cycle,右边有一个cycle,像∞这样,(或者上下各有个cycle,那就像8这样)。那么如果找到一个cycle后就把visit[]置1,那第二个cycle就没法被发现了。

  public List<List<Integer>> allCycles(Map<Integer, Set<Integer>> graph) {
    List<List<Integer>> result = new ArrayList<>();
    if (graph == null || graph.isEmpty()) {
      return result;
    }
 
    int[] visited = new int[graph.size()];
    for (int v : graph.keySet()) {
      if (visited[v] == 0) {
        List<Integer> tmp = new ArrayList<>();
        dfs(graph, v, visited, result, tmp);
      }
    }
    return result;
  }
 
  private void dfs(Map<Integer, Set<Integer>> graph, int v, int[] visited, List<List<Integer>> result, List<Integer> tmp) {
    if (visited[v] == -1) {
      List<Integer> cycle = new LinkedList<>();
      cycle.add(v);
      for (int i = tmp.size() - 1; i >= 0 ; i--) {
        cycle.add(0, tmp.get(i));
        if (tmp.get(i) == v) {
          break;
        }
      }
 
      result.add(new ArrayList<>(cycle));
      return;
    }
    if (visited[v] == 1) {
      return;
    }
 
    visited[v] = -1;
    tmp.add(v);
    for (int adj : graph.get(v)) {
      dfs(graph, adj, visited, result, tmp);
    }
    tmp.remove(tmp.size() - 1);
    visited[v] = 1;
  }
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