{"id":2413,"date":"2018-04-02T23:11:40","date_gmt":"2018-04-03T06:11:40","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=2413"},"modified":"2018-04-02T23:14:08","modified_gmt":"2018-04-03T06:14:08","slug":"leetcode-599-minimum-index-sum-of-two-lists","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-599-minimum-index-sum-of-two-lists\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 599. Minimum Index Sum of Two Lists"},"content":{"rendered":"

Problem<\/h1>\n

Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.<\/p>\n

You need to help them find out their\u00a0common interest<\/b>\u00a0with the\u00a0least list index sum<\/b>. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.<\/p>\n

Example 1:<\/b><\/p>\n

Input:<\/b>\r\n[\"Shogun\", \"Tapioca Express\", \"Burger King\", \"KFC\"]\r\n[\"Piatti\", \"The Grill at Torrey Pines\", \"Hungry Hunter Steakhouse\", \"Shogun\"]\r\nOutput:<\/b> [\"Shogun\"]\r\nExplanation:<\/b> The only restaurant they both like is \"Shogun\".\r\n<\/pre>\n
Example 2:<\/b><\/p>\n
Input:<\/b>\r\n[\"Shogun\", \"Tapioca Express\", \"Burger King\", \"KFC\"]\r\n[\"KFC\", \"Shogun\", \"Burger King\"]\r\nOutput:<\/b> [\"Shogun\"]\r\nExplanation:<\/b> The restaurant they both like and have the least index sum is \"Shogun\" with index sum 1 (0+1).\r\n<\/pre>\n
Note:<\/b><\/p>\n
    \n
  1. The length of both lists will be in the range of [1, 1000].<\/li>\n
  2. The length of strings in both lists will be in the range of [1, 30].<\/li>\n
  3. The index is starting from 0 to the list length minus 1.<\/li>\n
  4. No duplicates in both lists.<\/li>\n<\/ol>\n
    Solution<\/strong><\/h1>\n
    Time complexity: O(n+m)<\/p>\n
    Space complexity: O(n)<\/p>\n
    C++<\/p>\n
    \/\/ Author: Huahua\r\n\/\/ Running time: 103 ms\r\nclass Solution {\r\npublic:\r\n  vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {\r\n    unordered_map<string, int> indices;\r\n    for (int i = 0; i < list1.size(); ++i)\r\n      indices[list1[i]] = i;\r\n    \r\n    vector<string> ans; \r\n    int best_index = INT_MAX;\r\n    string best;\r\n    \r\n    for (int i = 0; i < list2.size(); ++i) {\r\n      if (!indices.count(list2[i])) continue;\r\n      int index = indices[list2[i]] + i;\r\n      if (index < best_index) ans.clear(); \r\n      if (index <= best_index) {\r\n        best_index = index;\r\n        ans.push_back(list2[i]);\r\n      }\r\n    }    \r\n    \r\n    return ans;\r\n  }\r\n};<\/pre>\n
     <\/p>\n","protected":false},"excerpt":{"rendered":"
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