【LeetCode】599. 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.

代码实现:

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string,int> m;
        unordered_map<string, int>::iterator iter;
        vector<string> ret;
        int cnt;
        int min=INT_MAX;
        
        for(int i=0;i<list1.size();i++){
            m.insert(pair<string, int>(list1[i],i));
        }
        for(int j=0;j<list2.size();j++){
            iter = m.find(list2[j]);
            if(iter != m.end()){
                cnt = iter->second + j;
                if(min>cnt){
                    min = cnt;
                    if(!ret.empty()){
                        ret.clear();
                    }
                    ret.push_back(iter->first);
                }else if(min==cnt){
                    ret.push_back(iter->first);
                }
            }
        }
        return ret;
    }
};


 
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