[Leetcode] 599. Minimum Index Sum of Two Lists 解题报告

题目：

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.
思路：

基本思路是：1）对于list2中的每个string，在list1中进行查找，如果找到了，就计算它们的index之和，并记录下来。最后返回index之和最小的所有string即可。

用暴力搜索法的话时间复杂度是O(mn)，其中m和n分别是list1和list2的长度。但是我们可以首先建立一个hash1，为list1中的string到其index的映射（这样就可以在O(1)的时间复杂度内找到list1中的某个string），然后建立一个从int到vector<string>的映射hash2，一旦在list1中找到了list2中的某个单词，就将其加入该映射。最后返回hash2中key最小的值即可。这样时间复杂度就降低到O(n)了，而且只需要两遍扫描。

代码：

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string, int> hash1;
        for (int i = 0; i < list1.size(); ++i) {
            hash1[list1[i]] = i;
        }
        unordered_map<int, vector<string>> hash2;
        int min_index_sum = INT_MAX;
        for (int i = 0; i < list2.size(); ++i) {
            auto it = hash1.find(list2[i]);
            if (it != hash1.end()) {
                hash2[i + it->second].push_back(list2[i]);
                min_index_sum = min(min_index_sum, i + it->second);
            }
        }
        return hash2[min_index_sum];
    }
};