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.
class Solution { public String[] findRestaurant(String[] list1, String[] list2) { List<String> res = new ArrayList(); List<String> l2 = Arrays.asList(list2); int min = 2000; for(int i = 0; i < list1.length; i++) { int ind = l2.indexOf(list1[i]); if(ind >= 0) { if(ind + i < min) { res.clear(); res.add(list1[i]); min = ind + i; } else if(ind + i == min) res.add(list1[i]); } } String[] r = new String[res.size()]; int i = 0; for(String s: res) r[i++] = s; return r; } }
brute force
public class Solution { public String[] findRestaurant(String[] list1, String[] list2) { HashMap < String, Integer > map = new HashMap < String, Integer > (); for (int i = 0; i < list1.length; i++) map.put(list1[i], i); List < String > res = new ArrayList < > (); int min_sum = 4000; int sum = 0; for (int j = 0; j < list2.length; j++) { if (map.containsKey(list2[j])) { sum = j + map.get(list2[j]); if (sum < min_sum) { res.clear(); res.add(list2[j]); min_sum = sum; } else if (sum == min_sum) res.add(list2[j]); } } return res.toArray(new String[res.size()]); } }
用hashmap,O(l1 + l2)