Minimum Index Sum of Two Lists问题及解法

问题描述：

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.

示例：

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".

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).

问题分析：

遍历两个数组，统计相同元素的索引和，始终将索引和最小的保存下来。最后，根据索引求元素值。

过程详见代码：

class Solution {public:    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {        vector<vector<pair<int, int>>> res(1),temp(1);int mini = list1.size() + list2.size();for (int i = 0; i < list1.size(); i++){for (int j = 0; j < list2.size(); j++){if (list1[i] == list2[j]){if (i + j < mini){mini = i + j;temp = res;temp[0].push_back(pair<int, int>(i, j));}else if (i + j == mini){temp[0].push_back(pair<int, int>(i, j));}}}}vector<string> vres;for (int i = 0; i < temp[0].size(); i++){vres.push_back(list1[temp[0][i].first]);}return vres;    }};