[leetcode-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:
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：

主要是用map记录字符串出现次数，以及记录字符串对应的下标。

感觉写的太啰嗦，待优化。

vector<string> findRestaurant(vector<string>& list1, vector<string>& list2)
{
  vector<string>res;
  map<string,int>restaurant;//记录是否 饭店出现次数。如果>1
  map<string,int>index1;//记录饭店1索引
  map<string,int>index2;//记录饭店2索引
  map<string,int>::iterator it;
  for(int i=;i<list1.size();i++)
  {
    index1[list1[i]] = i;
    restaurant[list1[i]]++;
  }
  for(int i=;i<list2.size();i++)
  {
    index2[list2[i]] = i;
    restaurant[list2[i]]++;
  }
  int indexsum = ;
  for(it =restaurant.begin();it!=restaurant.end();it++)
  {
     if(it->second ==)
     {
       if(index1[it->first] + index2[it->first] <indexsum)
       {
     indexsum =index1 [it->first] + index2[it->first];
     res.clear();
     res.push_back(it->first);
      }
      else if(index1[it->first] + index2[it->first] == indexsum)
      {
     indexsum =index1 [it->first] + index2[it->first];
     res.push_back(it->first);
      }
    }
  }
  return res;
}