[leetcode-599-Minimum Index Sum of Two Lists]

最新推荐文章于 2025-07-29 14:24:32 发布

转载最新推荐文章于 2025-07-29 14:24:32 发布 · 93 阅读

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原文链接：http://www.cnblogs.com/hellowooorld/p/6915305.html

本文介绍了一个帮助两个人基于各自的餐厅喜好列表找到共同喜欢的餐厅的算法。该算法通过使用映射记录每个餐厅出现的次数及其在列表中的位置来实现，最终找出双方共同喜爱且列表索引总和最小的餐厅。

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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:
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：

主要是用map记录字符串出现次数，以及记录字符串对应的下标。

感觉写的太啰嗦，待优化。

vector<string> findRestaurant(vector<string>& list1, vector<string>& list2)
{
  vector<string>res;
  map<string,int>restaurant;//记录是否 饭店出现次数。如果>1 
  map<string,int>index1;//记录饭店1索引
  map<string,int>index2;//记录饭店2索引
  map<string,int>::iterator it;
  for(int i=0;i<list1.size();i++)
  {
    index1[list1[i]] = i;
    restaurant[list1[i]]++;
  }
  for(int i=0;i<list2.size();i++)
  {
    index2[list2[i]] = i;
    restaurant[list2[i]]++;
  }
  int indexsum = 2000;  
  for(it =restaurant.begin();it!=restaurant.end();it++)
  {     
     if(it->second ==2) 
     {          
       if(index1[it->first] + index2[it->first] <indexsum)
       {
     indexsum =index1 [it->first] + index2[it->first];
     res.clear();
     res.push_back(it->first);
      }
      else if(index1[it->first] + index2[it->first] == indexsum)
      {
     indexsum =index1 [it->first] + index2[it->first];     
     res.push_back(it->first);
      }       
    }     
  }    
  return res;
}

转载于:https://www.cnblogs.com/hellowooorld/p/6915305.html