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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.

思路：此题用到了两个map映射，第一个map res映射第一个vector中的string及其相对应的位置，并在第二个vector中顺序查找是否出现在 res 中，若存在，则判定此string 的位置和对应res中对应string的位置之和是否为小于当前的 ll，若小于则添加进去。

代码如下：

vector
   
     findRestaurant(vector
    
     & list1, vector
     
      & list2) {    vector
      
        res;    if (list1.empty() || list2.empty())return res;    multimap
       
         data;    unordered_map
        
          temp; for (int i = 0; i < list1.size(); i++)temp[list1[i]] = i; int ll = INT_MAX; for (int i = 0; i < list2.size(); i++){ if (temp.find(list2[i]) != temp.end()){ int tt = temp[list2[i]] + i; if (tt <= ll){ data.insert({ tt, list2[i] }); ll = tt; } } } auto it = data.begin(); int len = it->first; for (; it != data.end(); it++){ if (it->first == len)res.push_back(it->second); else break; } return res;}
        
       
      
     
    
   

算法的时间复杂度为O(n)，空间复杂度为O(m+n)，m,n分别为两个vector的长度。

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