Problem:
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.
思路:
Solution (C++):
vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
int m = list1.size(), n = list2.size();
int min_index = INT_MAX;
vector<string> res;
unordered_map<string, int> hash;
for (int i = 0; i < m; ++i) {
hash.emplace(list1[i], i);
}
for (int i = 0; i < n; ++i) {
if (hash.count(list2[i]) == 0) continue;
int sum = i + hash[list2[i]];
if (sum < min_index) res.clear();
if (sum <= min_index) {
res.push_back(list2[i]);
min_index = sum;
}
}
return res;
}
性能:
Runtime: 120 ms Memory Usage: 25.8 MB
思路:
Solution (C++):
性能:
Runtime: ms Memory Usage: MB