QUESTION:
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).
Note:
- The length of both lists will be in the range of [1, 1000].
- The length of strings in both lists will be in the range of [1, 30].
- The index is starting from 0 to the list length minus 1.
- No duplicates in both lists.
EXPLANATION:
其实这道题就是按部就班的来,先将list1的数据都存入,然后再用list2的数据进行比对,如果相等,同时index+i的值也小于上次标记的值,那么就可以替换结果。
这里我觉得可能会出现0-5,5-0这种情况,也就是会有两种答案或者3种答案的情况,所以我在此处添加了==的情况。
SOLUTION:
public class Solution {
public String[] findRestaurant(String[] list1, String[] list2) {
int value = Integer.MAX_VALUE;
HashSet<String> result = new HashSet<>();
Hashtable<String,Integer> table = new Hashtable<>();
for(int i = 0;i<list1.length;i++){
String tmp = list1[i];
table.put(tmp,i);
}
for(int i = 0;i<list2.length;i++){
String tmp = list2[i];
int index = table.getOrDefault(tmp,-1);
if(index != -1 && index+i<value){
result.clear();
result.add(tmp);
value = index+i;
}
if(index != -1 && index+i==value){
result.add(tmp);
}
}
String[] stringResult = new String[result.size()];
Iterator<String> iterator = result.iterator();
for(int i = 0;i<result.size();i++){
stringResult[i] = iterator.next();
}
return stringResult;
}
}