1.排序

Collections.sort(list,(o1, o2)-> o1.get(0).compareTo(o2.get(0)));

2.返回值

3.往集合添加元素

Arrays.asList(元素)

 List<List<String>> list = new ArrayList<>();
        List<String> path = new ArrayList<>();

// 将[["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]加入list中
        List<String> entry1 = new ArrayList<>(Arrays.asList("JFK", "SFO"));
        List<String> entry2 = new ArrayList<>(Arrays.asList("JFK", "ATL"));
        List<String> entry3 = new ArrayList<>(Arrays.asList("SFO", "ATL"));
        List<String> entry4 = new ArrayList<>(Arrays.asList("ATL", "JFK"));
        List<String> entry5 = new ArrayList<>(Arrays.asList("ATL", "SFO"));

        list.add(entry1);
        list.add(entry2);
        list.add(entry3);
        list.add(entry4);
        list.add(entry5);

4.path为路线经过机场数，list为机票数，四张机票会经过5个机场

if (path.size() == tickets.size() + 1) {
            res = new LinkedList(path);
            return true;
        }

5.整体代码

class Solution {
    private LinkedList<String> res;
    private LinkedList<String> path = new LinkedList<>();

    public List<String> findItinerary(List<List<String>> tickets) {
        Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));
        path.add("JFK");
        boolean[] used = new boolean[tickets.size()];
        backTracking((ArrayList) tickets, used);
        return res;
    }

    public boolean backTracking(ArrayList<List<String>> tickets, boolean[] used) {
        if (path.size() == tickets.size() + 1) {
            res = new LinkedList(path);
            return true;
        }

        for (int i = 0; i < tickets.size(); i++) {
            if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) {
                path.add(tickets.get(i).get(1));
                used[i] = true;

                if (backTracking(tickets, used)) {
                    return true;
                }

                used[i] = false;
                path.removeLast();
            }
        }
        return false;
    }
}