LeetCode-鑰匙和房間

首先嚐試用BFS算法，遍歷到不同的鑰匙，總房間數減一，最終房間數等於0，就是全部能進去

public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        if (rooms == null || rooms.isEmpty()){
            return false;
        }

        int roomNum = rooms.size();

        if(roomNum == 1){
            return true;
        }

        List<Integer> keys = rooms.get(0);
        if (keys == null || keys.size() <= 0){
            return false;
        }

        Set<Integer> set = new HashSet<>();
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(0);
        while (!queue.isEmpty() && roomNum != 0){
            Integer poll = queue.poll();
            if (set.contains(poll)){
                continue;
            }else{
                roomNum --;
                set.add(poll);
                List<Integer> keyList = rooms.get(poll);
                if (keyList != null && keyList.size() > 0){
                    keyList.forEach(k -> queue.offer(k));
                }
            }
        }
        if (roomNum == 0){
            return true;
        }
        return false;
    }

因爲有限條路就能走遍所有房間的概率較大，所以DFS是要快一點的

private int roomNum;

    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        if (rooms == null || rooms.isEmpty()){
            return false;
        }

        roomNum = rooms.size();

        if(roomNum == 1){
            return true;
        }

        boolean[] already = new boolean[roomNum];

        dfs(rooms,already,0);

        if (roomNum == 0){
            return true;
        }

        return false;
    }

    private void dfs(List<List<Integer>> rooms, boolean[] already, int key){
        if (already[key]){
            return;
        }
        roomNum --;
        already[key] = true;
        if (roomNum == 0){
            return;
        }
        List<Integer> integers = rooms.get(key);
        if (integers.size() <= 0){
            return;
        }
        for (int room : rooms.get(key)){
            dfs(rooms, already, room);
        }
    }