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 = ; i < list1.size(); i++){ // 存入hashtable中
data.insert(pair<string, int>(list1[i], i));
}
for (int i = ; 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;
}
};