[Leetcode] 802. Find Eventual Safe States 解题报告_802. find eventual safe states解题报告 python

题目：

代码：

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.

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].

思路：

这道题目的实质就是找不在环上的节点。我们采用染色的方法对节点进行分类：0表示该结点还没有被访问；1表示已经被访问过了，并且发现是safe的；2表示被访问过了，但发现是unsafe的。我们采用DFS的方法进行遍历，并返回该结点是否是safe的：如果发现它已经被访问过了，则直接返回是否是safe的标记；否则就首先将其标记为unsafe的，然后进行DFS搜索（此时该结点会处在DFS的路径上，所以后面的DFS一旦到了该结点，就会被认为是形成了环，所以直接返回false）。当整个DFS的搜索都已经结束，并且都没有发现该结点处在环上时，说明该结点是safe的，所以此时将其最终标记为safe即可。

整个算法的空间复杂度是O(n)，时间复杂度是O(n)，因为每个结点最多会被访问一次。

代码：

class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        vector<int> res;
        if (graph.size() == 0) {
            return res;
        }
        int size = graph.size();
        vector<int> color(size, 0);         // 0: not visited; 1: safe; 2: unsafe.
        for (int i = 0; i < size; ++i) {
            if (dfs(graph, i, color)) {     // the i-th node is safe
                res.push_back(i);
            }
        }
        return res;
    }
private:
    bool dfs(vector<vector<int>> &graph, int start, vector<int> &color) {
        if (color[start] != 0) {
            return color[start] == 1;
        }
        color[start] = 2;       // mark it as unsafe because it is on the path
        for (int next : graph[start]) {
            if (!dfs(graph, next, color)) {
                return false;
            }
        }
        color[start] = 1;       // mark it as safe because no loop is found
        return true;
    }
};