599. Minimum Index Sum of Two Lists

题目来源【Leetcode】

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.

找出两个列表的公共元素且元素下标最小

方法一：直接就循环的找，比较麻烦

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        vector<string>s;
        vector<pair<string,int>>re;
        for(int i = 0; i < list1.size(); i++){
            for(int j = 0; j < list2.size(); j++){
                if(list1[i] == list2[j]) {
                    pair<string,int>t;
                    t.first = list1[i];
                    t.second = i+j;
                    re.push_back(t);
                }
            }
        }
        int mi =  list1.size()+list2.size();
        for(int i = 0;  i < re.size(); i++){
            mi = min(mi, re[i].second);
        }
        for(int i = 0; i < re.size(); i++){
            if(re[i].second == mi) s.push_back(re[i].first);
        }

        return s;
    }
};

第二种方法：可以用undered_map来做：

class Solution {
public:
vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
unordered_map< string, int> m;
vector<string> res;
int min_index_sum = 999999;
for(int i = 0; i < list1.size(); ++i) {
m[list1[i]] = i;
}
for(int i = 0; i < list2.size(); ++i) {
if(m.count(list2[i])) {
int s = i + m[list2[i]];
if(s == min_index_sum) {
res.push_back(list2[i]);
}
if(s < min_index_sum) {
res.clear();
res.push_back(list2[i]);
min_index_sum = s;
}
}
}
return res;
}
};