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;}
}