802. Find Eventual Safe States

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：
题解：
其实就是让判断一个点是不是出于环中，输出所有无环的路径上的点。
于是很明显是dfs，并且可以用dp, 即map来存储中间结果

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看到一个很好的思路，我们可以把map和visited数组combine, 如果为2，表示有环（该次visited过），如果为1， 表示之前已经检查过，没有环

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;
    }
}