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