LeetCode Problems #802

2018年10月7日

#802. Find Eventual Safe States

问题描述：

In a directed graph, we start at some node and every turn, walk along a directed edge of the graph.  If we reach a node that is terminal (that is, it has no outgoing directed edges), we stop.

Now, say our starting node is eventually safe if and only if we must eventually walk to a terminal node.  More specifically, there exists a natural number K so that for any choice of where to walk, we must have stopped at a terminal node in less than K steps.

Which nodes are eventually safe?  Return them as an array in sorted order.

The directed graph has N nodes with labels 0, 1, ..., N-1, where N is the length of graph.  The graph is given in the following form: graph[i] is a list of labels j such that (i, j) is a directed edge of the graph.

样例:

Example:
Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
Output: [2,4,5,6]
Here is a diagram of the above graph.

Note:

• graph will have length at most 10000.
• The number of edges in the graph will not exceed 32000.
• Each graph[i] will be a sorted list of different integers, chosen within the range [0, graph.length - 1].

问题分析：

本题难度为Medium！已给出的函数定义为：

class Solution:
    def eventualSafeNodes(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: List[int]
        """

其中graph为一个二维数组，返回值为一个列表；

算法如下：

使用深度优先查找，查找是否存在结点进入循环，即是否会访问到已经访问过的结点，若存在，则不是安全点，否则是。

代码如下：

#coding=utf-8
class Solution:
    def eventualSafeNodes(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: List[int]
        """
        def dfs(seen, path, isSafe):
            cur = path.pop(0) #取出结点

            for nextNode in graph[cur]:
                if nextNode in seen:
                    isSafe[0]=False
                    return
                cp_path=path.copy() #添加结点
                cp_path.append(nextNode)
                cp_seen=seen.copy() #标记访问结点
                cp_seen.append(nextNode)
                dfs(cp_seen,cp_path,isSafe)
        
        res=[]
        for node in range(len(graph)):
            seen=[node]
            path=[node]
            isSafe=[True]
            dfs(seen,path,isSafe)
            if isSafe[0]==True:
                res.append(node)
        return res

以上算法可能会出现时间限制问题。