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.
/* 暴力的话O(n^2),可以利用hashtable,用空间来换时间。 */ class Solution { public: vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) { vector<string> res; map<string, int> data; int minval = INT_MAX; for (int i = 0; i < list1.size(); i++){ // 存入hashtable中 data.insert(pair<string, int>(list1[i], i)); } for (int i = 0; i < list2.size(); i++){ if (data.find(list2[i]) != data.end()){ if (data[list2[i]] + i < minval){ minval = data[list2[i]] + i; res.clear(); res.push_back(list2[i]); }else if (data[list2[i]] + i == minval){ res.push_back(list2[i]); } } } return res; } };