# 题目描述：

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.

# 代码实现：

``````class Solution {
public:
vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
unordered_map<string,int> m;
unordered_map<string, int>::iterator iter;
vector<string> ret;
int cnt;
int min=INT_MAX;

for(int i=0;i<list1.size();i++){
m.insert(pair<string, int>(list1[i],i));
}
for(int j=0;j<list2.size();j++){
iter = m.find(list2[j]);
if(iter != m.end()){
cnt = iter->second + j;
if(min>cnt){
min = cnt;
if(!ret.empty()){
ret.clear();
}
ret.push_back(iter->first);
}else if(min==cnt){
ret.push_back(iter->first);
}
}
}
return ret;
}
};
``````