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.

這道題給了我們兩個字串陣列，讓我們找到座標位置之和最小的相同的字串。那麼對於這種陣列項和其座標之間關係的題，最先考慮到的就是要建立資料和其位置座標之間的對映。我們建立list1的值和座標的之間的對映，然後遍歷list2，如果當前遍歷到的字串在list1中也出現了，那麼我們計算兩個的座標之和，如果跟我們維護的最小座標和mn相同，那麼將這個字串加入結果res中，如果比mn小，那麼mn更新為這個較小值，然後將結果res清空並加入這個字串，參見程式碼如下：

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        vector<string> res;
        unordered_map<string, int> m;
        int mn = INT_MAX, n1 = list1.size(), n2 = list2.size();
        for (int i = 0; i < n1; ++i) m[list1[i]] = i;
        for (int i = 0; i < n2; ++i) {
            if (m.count(list2[i])) {
                int sum = i + m[list2[i]];
                if (sum == mn) res.push_back(list2[i]);
                else if (sum < mn) {
                    mn = sum;
                    res = {list2[i]};
                }
            }
        }
        return res;
    }
};

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