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.

Illustration of 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].

一刷：
題解：
其實就是讓判斷一個點是不是出于環(huán)中，輸出所有無環(huán)的路徑上的點。
于是很明顯是dfs，并且可以用dp, 即map來存儲中間結(jié)果

class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
         Map<Integer, Boolean> map = new HashMap<>();
         int len = graph.length;
         
        
        for(int i = 0; i<graph.length; i++){
            boolean[] visited = new boolean[len];
            if(map.containsKey(i)) continue;
            boolean status = true;
            int[] neighbor = graph[i];
            if(neighbor.length == 0) map.put(i, true);
            for(int n : neighbor){
                status = status && dfs(n, map, graph, visited);
            }
            map.put(i, status);
        }
        
        List<Integer> res = new ArrayList<>();
        for (Map.Entry<Integer, Boolean> entry : map.entrySet())
        {
            if(entry.getValue()) res.add(entry.getKey());
        }
        return res;
    }
    
    private boolean dfs(int i, Map<Integer, Boolean> map, int[][] graph, boolean[] visited){
        if(map.containsKey(i)) return map.get(i);
        if(visited[i]){
            map.put(i, false);
            return false;
        }
        visited[i] = true;
        
        int[] neighbor = graph[i];
        if(neighbor.length == 0){
            map.put(i, true);
            return true;
        }
        boolean status = true;
        for(int node : neighbor){
            status = status && dfs(node, map, graph, visited);
        }
        map.put(i, status);
        return status;
    }
}

在discuss看到一個很好的思路，我們可以把map和visited數(shù)組combine, 如果為2，表示有環(huán)（該次visited過），如果為1， 表示之前已經(jīng)檢查過，沒有環(huán)

class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        List<Integer> res = new ArrayList<>();
        if(graph == null || graph.length == 0) return res;
        
        int nodeCount = graph.length;
        int[] color = new int[nodeCount];
        for(int i=0; i<nodeCount; i++){
            if(dfs(graph, i, color)) res.add(i);
        }
        return res;
    }
    
    private boolean dfs(int[][] graph, int start, int[] color){
        if(color[start] != 0)   return color[start] == 1;//1->true, 2->false
        color[start] = 2;
        for(int neighbor : graph[start]){
            if(!dfs(graph, neighbor, color)) return false;
        }
        color[start] = 1;
        return true;
    }
}