LeetCode599. Minimum Index Sum of Two Lists [C++]

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".

解题思路：

1、创建map，以店名为键，索引和为值。为判断哪些为两次，每次再减sum为值。

2、寻找map中最小的值。

3、找到最小值所对应的键（有可能不止一个）。

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        map<string,int> m;
        int sum=list1.size()+list2.size();
        for(int i=0;i<list1.size();i++)
        {
            m[list1[i]]=m[list1[i]]+i-sum;
        }
        for(int i=0;i<list2.size();i++)
        {
            m[list2[i]]=m[list2[i]]+i-sum;
        }
        int min=0;
        for(map<string,int>::iterator it=m.begin();it!=m.end();it++)
        {
            if(it->second<min)
                min=it->second;
            //cout<<it->first<<" "<<it->second<<endl;
        }
        vector<string> ans;
        for(map<string,int>::iterator it=m.begin();it!=m.end();it++)
        {
            if(it->second==min)
            {
                ans.push_back(it->first);
            }
        }
        return ans;
    }
};