599. Minimum Index Sum of Two Lists

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:
1. The length of both lists will be in the range of [1, 1000].
2. The length of strings in both lists will be in the range of [1, 30].
3. The index is starting from 0 to the list length minus 1.
4. No duplicates in both lists.

思路：此题用到了两个map映射，第一个map res映射第一个vector中的string及其相对应的位置，并在第二个vector中顺序查找是否出现在 res 中，若存在，则判定此string 的位置和对应res中对应string的位置之和是否为小于当前的 ll，若小于则添加进去。
代码如下：

vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {    vector<string> res;    if (list1.empty() || list2.empty())return res;    multimap<int, string> data;    unordered_map<string, int> temp;    for (int i = 0; i < list1.size(); i++)temp[list1[i]] = i;    int ll = INT_MAX;    for (int i = 0; i < list2.size(); i++){        if (temp.find(list2[i]) != temp.end()){            int tt = temp[list2[i]] + i;            if (tt <= ll){                data.insert({ tt, list2[i] });                ll = tt;            }        }    }    auto it = data.begin();    int len = it->first;    for (; it != data.end(); it++){        if (it->first == len)res.push_back(it->second);        else break;    }    return res;}

算法的时间复杂度为O(n)，空间复杂度为O(m+n)，m,n分别为两个vector的长度。