599. Minimum Index Sum of Two Lists -E

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.

A:

--------------------就是用个map以后，逐个对比

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string, int> m1;
        for(int i =0;i<list1.size();++i)
            m1.insert({list1[i],i});
        
        int minSum = list1.size() + list2.size();
        vector<string> res;
        for(int i =0;i<list2.size();++i)
        {
            auto it = m1.find(list2[i]);
            if( it != m1.end())
            {
                if(minSum > i+ it->second)
                {
                    minSum = i + it->second;
                    res = {list2[i]};
                }else if(minSum == i + it->second)
                {
                    res.push_back(list2[i]);
                }
            }
        }
        return res;           
    }
};