599. Minimum Index Sum of Two Lists

Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.

You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.

Example 1:

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
Output: ["Shogun"]
Explanation: The only restaurant they both like is "Shogun".

Example 2:

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["KFC", "Shogun", "Burger King"]
Output: ["Shogun"]
Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).


输出两个字符串数组共有的字符串，要求索引之和最小

C++(92ms):

 1 class Solution {
 2 public:
 3     vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
 4         vector<string> res ;
 5         unordered_map<string,int> Map ;
 6         int Min = INT_MAX ;
 7         for(int i = 0 ; i < list1.size() ; i++){
 8             Map[list1[i]] = i ;
 9         }
10         for(int i = 0 ; i < list2.size() ; i++){
11             if (Map.count(list2[i])){
12                 if (Map[list2[i]] + i < Min ){
13                     Min = Map[list2[i]] + i ;
14                     res.clear();
15                     res.push_back(list2[i]) ;
16                 }else if(Map[list2[i]] + i == Min ){
17                     res.push_back(list2[i]) ;
18                 }
19             }
20         }
21         return res ;
22     }
23 };

Java(29ms):

 1 class Solution {
 2     public String[] findRestaurant(String[] list1, String[] list2) {
 3         HashMap<String,Integer> Map = new HashMap() ;
 4         LinkedList<String> res = new LinkedList() ;
 5         int Min = Integer.MAX_VALUE ;
 6         for(int i = 0 ; i < list1.length ; i++){
 7             Map.put(list1[i] , i) ;
 8         }
 9         for(int i = 0 ; i < list2.length ; i++){
10             if (Map.containsKey(list2[i])){
11                if (Map.get(list2[i]) + i < Min){
12                    Min = Map.get(list2[i]) + i ;
13                    res.clear() ;
14                    res.add(list2[i]) ;
15                }else if (Map.get(list2[i]) + i == Min){
16                    res.add(list2[i]) ;
17                }
18             }
19         }
20         return res.toArray(new String[res.size()]) ;
21     }
22 }