599. Minimum Index Sum of Two Lists*

599. Minimum Index Sum of Two Lists*

​​https://leetcode.com/problems/minimum-index-sum-of-two-lists/​​

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.

C++ 实现 1

使用 ​​record​​​ 这个哈希表来记录重合的字符串以及对应的索引之和, 在遍历 ​​record​​ 查找 index sum 最小的结果.

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string, int> tmp, record;
        for (int i = 0; i < list1.size(); ++ i) tmp[list1[i]] = i;
        for (int i = 0; i < list2.size(); ++ i) {
            if (tmp.count(list2[i]))
                record[list2[i]] = i + tmp[list2[i]];
        }
        int k = INT32_MAX;
        vector<string> res;
        for (auto &p : record) {
            if (p.second == k)
                res.push_back(p.first);
            else if (p.second < k) {
                res = {p.first};
                k = p.second;
            }
        }
        return res;
    }
};

C++ 实现 2

下面代码只用一个哈希表, 但逻辑稍复杂, 如果查找到重合的字符串, 就将索引之和表示为负数. 由于重合元素的索引之和为负数, 所以之后要找负数中最大的那些结果.

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string, int> record;
        for (int i = 0; i < list1.size(); ++i)
            record[list1[i]] = i + 1;

        for (int i = 0; i < list2.size(); ++i) {
            if (record.count(list2[i])) {
                record[list2[i]] = -record[list2[i]];
                record[list2[i]] += -(i + 1);
            }
        }

        vector<string> res;
        int imax = INT32_MIN;
        for (auto &iter : record) {
            if (iter.second < 0) {
                if (iter.second == imax)
                    res.push_back(iter.first);
                else if (iter.second > imax) {
                    imax = iter.second;
                    res = vector<string>{iter.first};
                }
            }
        }
        return res;
    }
};