1436. Destination City
You are given the array paths
, where paths[i] = [cityAi, cityBi]
means there exists a direct path going from cityAi
to cityBi
. Return the destination city, that is, the city without any path outgoing to another city.
It is guaranteed that the graph of paths forms a line without any loop, therefore, there will be exactly one destination city.
Example 1:
Input: paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]] Output: "Sao Paulo" Explanation: Starting at "London" city you will reach "Sao Paulo" city which is the destination city. Your trip consist of: "London" -> "New York" -> "Lima" -> "Sao Paulo".
Example 2:
Input: paths = [["B","C"],["D","B"],["C","A"]] Output: "A" Explanation: All possible trips are: "D" -> "B" -> "C" -> "A". "B" -> "C" -> "A". "C" -> "A". "A". Clearly the destination city is "A".
Example 3:
Input: paths = [["A","Z"]] Output: "Z"
Constraints:
1 <= paths.length <= 100
paths[i].length == 2
1 <= cityAi.length, cityBi.length <= 10
cityAi != cityBi
- All strings consist of lowercase and uppercase English letters and the space character.
class Solution { public String destCity(List<List<String>> paths) { List<String> list = new ArrayList(); for(int i = 0; i < paths.size(); i++){ if(!list.contains(paths.get(i).get(1))) list.add(paths.get(i).get(1)); else list.remove(paths.get(i).get(1)); } for(int i = 0; i < paths.size(); i++){ if(list.contains(paths.get(i).get(0))) list.remove(paths.get(i).get(0)); } return list.get(0); } }
观察,得到特征:所有destination都出现一次且只出现在list的第二位,第一遍把独特的list第二个city找出来,第二遍遍历把独一无二的找出来就可以了
class Solution { public String destCity(List<List<String>> paths) { Set<String> set= new HashSet<>(); for (List<String> l: paths) set.add(l.get(1)); for (List<String> l: paths) set.remove(l.get(0)); return set.iterator().next(); } }
这种方法更好